11. If v,w,x,y,z are non negative integers each less than 11, then how many distinct combinations are possible of(v,w,x,y,z) which satisfy v(11^4) +w(11^3)+ x(11^2)+ y(11) +z = 151001 ?
Changing 151001 to base 11 number we get,
it will be
A34A4 i.e.
10 3 4 10 4 v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001 where v=10, w=3,x=4, y=10, z=4
it will be
A34A4 i.e.
10 3 4 10 4 v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001 where v=10, w=3,x=4, y=10, z=4
12. In a certain examination paper there are n questions. For j=1,2,3,.....n, there are 2^(n-1) students whoanswered j or more question wrongly. If the total number of wrong answers is 4096 then the value of n is
a)13
b)11
c)10
d)9
Given that,
2^(n-1) = 4096 = 2^12
i.e)n-1=12
==> n=13
2^(n-1) = 4096 = 2^12
i.e)n-1=12
==> n=13
13. How many six digit number can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at it's unit's place?
As each number will contain all the six digits and the sum of digits is = 1+2+3+4+5+6 = 21 which is divisible by 3. So each number is divisible by 3.
The numbers ending with digit 1 will be divisible 1.
The numbers ending with digit 2 will be divisible 2.
The numbers ending with digit 3 are divisible 3.
The numbers ending with digit 5 will be divisible 5.
The numbers ending with digit 6 will be divisible 6.
Except the numbers ending with last two digits as 14, 34 and 54 all other numbers ending with 4 are divisible by 4.
The no. of numbers ending with last two digits 14,34 and 54 are = 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4)
so six digit number that can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place = 720-72= 648
The numbers ending with digit 1 will be divisible 1.
The numbers ending with digit 2 will be divisible 2.
The numbers ending with digit 3 are divisible 3.
The numbers ending with digit 5 will be divisible 5.
The numbers ending with digit 6 will be divisible 6.
Except the numbers ending with last two digits as 14, 34 and 54 all other numbers ending with 4 are divisible by 4.
The no. of numbers ending with last two digits 14,34 and 54 are = 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4)
so six digit number that can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place = 720-72= 648
14. A natural number has exactly 10 divisors including 1 and itself. how many distinct prime factors can this natural number can have?
Consider the no is 512, the divisors of this no. are
1,2,4,8,16,32,64,128,256,512;
so only one prime factors are there i.e. 2;
if the no is 48 the divisors are
1,2,3,4,6,8,12,16,24,48;
so the prime factors are 2,3;
1,2,4,8,16,32,64,128,256,512;
so only one prime factors are there i.e. 2;
if the no is 48 the divisors are
1,2,3,4,6,8,12,16,24,48;
so the prime factors are 2,3;
15. If m and n are two positive integers, then what is the value of mn? Given:
(1)7m + 5n= 29
(2) m + n= 5
7m + 5n = 29
m + n = 5(multiply both sides by 7)
subtract equation second from first,we get
7m + 5n =29
7m + 7n = 35
n = 3
substitute value of n in any equation.
m = 2
mn = 6.
m + n = 5(multiply both sides by 7)
subtract equation second from first,we get
7m + 5n =29
7m + 7n = 35
n = 3
substitute value of n in any equation.
m = 2
mn = 6.
16. A natural number has exactly 10 divisors including 1 and itself. How many distinct prime factors can this natural number have?
A. Either 1 or 2
B. Either 1 or 3
C. Either 2 or 3
D. Either 1,2 or 3
Solution:
Ans) Either 1 or 2
Check on 29 i.e 512 , 39, 59 which have only 1 prime factor and 80 , 48 which are having 2
prime factors and total of 10 divisors.
Let us also consider the case of 3 prime factors. Let x, y ,z be the three prime factors of a
number. Therefore 1 , x ,y ,z ,xy ,yx, zx ,xyz must be the factors of that nos . We have
minimum 8 such factors with xyz as the nos or the factor of the nos.
When xyz is the nos then we will have exactly 8 divisors but if the nos is greater thn xyz that
is a multiple of xyz , either the nos is multiplied by any of these prime factors x , y , z then
we will get at least 12 divisors. So we don’t get 3 prime factors with 10 divisors.
17. What is the remainder when 128^1000 is divided by 153?
128^1000 = (153-25)^1000 = (25^1000)mod153 = (625^500)mod153 = [(4*153+13)^500]mod153 = (13^500)mod153 = (169^250)mod153 = [(153+16)^250]mod153 = (16^250)mod153 = (256^125)mod153 = [(153+103)^125]mod153 = (103^125)mod153 = [(2*3*17+1)^125]mod153
At this point,observe that 153=17*(3^2);
Now,therefore clearly (2*3*17+1)^125 = [(125C124)*{(2*3*17)^1}*(1^124)+1]mod153.
Actually,the above line can be written since only except the last two
terms,every term of the expansion of (2*3*17+1)^125 has [(2*3*17)^2],i.e,
[153*68] as one of its factors.
Now, [(125C124)*{(2*3*17)^1}*(1^124)+1]= 125*2*3*17 + 1;
[125*2*3*17 + 1]/(153) = [125*2*3*17 + 1]/(3^2*17) = (125*2*3*17)/(3^2*17) + 1/(3^2*17) = 250/3 + 1/(3^2*17) =83 + (1/3)+ 1/(3^2*17) = 83 + (52/153);
which means [125*2*3*17 + 1] = 83*153 + 52;
which again implies 128^1000 = [125*2*3*17 + 1]mod153 = {83*153 + 52}mod153 = 52mod153 ;
So, remainder is 52.
At this point,observe that 153=17*(3^2);
Now,therefore clearly (2*3*17+1)^125 = [(125C124)*{(2*3*17)^1}*(1^124)+1]mod153.
Actually,the above line can be written since only except the last two
terms,every term of the expansion of (2*3*17+1)^125 has [(2*3*17)^2],i.e,
[153*68] as one of its factors.
Now, [(125C124)*{(2*3*17)^1}*(1^124)+1]= 125*2*3*17 + 1;
[125*2*3*17 + 1]/(153) = [125*2*3*17 + 1]/(3^2*17) = (125*2*3*17)/(3^2*17) + 1/(3^2*17) = 250/3 + 1/(3^2*17) =83 + (1/3)+ 1/(3^2*17) = 83 + (52/153);
which means [125*2*3*17 + 1] = 83*153 + 52;
which again implies 128^1000 = [125*2*3*17 + 1]mod153 = {83*153 + 52}mod153 = 52mod153 ;
So, remainder is 52.
18. Given a,b,c are in GP and a < b < c. How many different different values of a, b, c satisfy (log(a) + log(b) + log(c) ) = 6?
log(abc)=6
abc=10^6 because of b^2=ac=100,
b^3=10^6,
b=10^2=100
and the combinations are
(2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500).
abc=10^6 because of b^2=ac=100,
b^3=10^6,
b=10^2=100
and the combinations are
(2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500).
19. What will be the remainder when expression 2^2+22^2+222^2+2222^2+....+22222...48times^2 is divided by 9?
First, let us consider a general case :
(222222222222.....{2 is repeated n times})^2
=[2(111111111111........{1 is repeated n times})]^2
=4(111111111111........{1 is repeated n times})^2
=4{(10)^(n-1) + (10)^(n-2) + (10)^(n-3) + (10)^(n-4) + .... +1}^2.......[(eqn1)]
Now, we know 9 divides (10^k -1) , since (10^k -1)=(999999....k times);
Where k is ofcourse a +ve integer. So,(10^k -1) is of the form (9A + 1);[A being a +ve integer].
Now, looking at the [(eqn1)] ,
we can find that here k =(n-1), (n-2), (n-3),....,1.
Now,for 10^(n-1) = 9*(a1) + 1;
10^(n-2) = 9*(a2) + 1;
10^(n-3) = 9*(a3) + 1;
10^(n-4) = 9*(a4) + 1;
.
.
.
.
10^1 =9 +1;
1 = 1
So,now in [(eqn1)] , we can write
(222222222222.....{2 is repeated n times})^2
=4(9*A +{1 + 1 + 1 + 1 + .........[1 is added n times]})^2
=4*(9*A + n)^2; here, A =(a1+a2+a3+a4+....+1),of course an integer.
=4*81*(A^2) + 4*2*(9*A)*n + 4*(n^2);...............[(eqn 2)]
Now this a general expression for (222222222222.....{2 is repeated n times})^2.
For the given problem,we can find that n=1,2,3,4,5,6,....48.
In the [(eqn 2)],since 1st two terms are already divisible by 9,so we only have
to consider the last term,i.e, 4*(n^2);
Summing up [4*(n^2)] for n=1,2,...48. we get
4{(1^2)+(2^2)+(3^2)+(4^2)+(5^2)+...(48^2)}
=4*[48*(48+1)*(2*48 +1)/6] ; using the formula[1^2 +2^2 +..+n^2=n(n+1)(2n+1)/6]
=4*8*49*97
=4(9- 1)(9*5 +4)(9*11 -2)
=9*B + 4(-1)*(4)*(-2); B an integer, to know its value is not important
=9*B + 32
=9*B + 9*3 + 5;
So, clearly the remainder will be 5.
(222222222222.....{2 is repeated n times})^2
=[2(111111111111........{1 is repeated n times})]^2
=4(111111111111........{1 is repeated n times})^2
=4{(10)^(n-1) + (10)^(n-2) + (10)^(n-3) + (10)^(n-4) + .... +1}^2.......[(eqn1)]
Now, we know 9 divides (10^k -1) , since (10^k -1)=(999999....k times);
Where k is ofcourse a +ve integer. So,(10^k -1) is of the form (9A + 1);[A being a +ve integer].
Now, looking at the [(eqn1)] ,
we can find that here k =(n-1), (n-2), (n-3),....,1.
Now,for 10^(n-1) = 9*(a1) + 1;
10^(n-2) = 9*(a2) + 1;
10^(n-3) = 9*(a3) + 1;
10^(n-4) = 9*(a4) + 1;
.
.
.
.
10^1 =9 +1;
1 = 1
So,now in [(eqn1)] , we can write
(222222222222.....{2 is repeated n times})^2
=4(9*A +{1 + 1 + 1 + 1 + .........[1 is added n times]})^2
=4*(9*A + n)^2; here, A =(a1+a2+a3+a4+....+1),of course an integer.
=4*81*(A^2) + 4*2*(9*A)*n + 4*(n^2);...............[(eqn 2)]
Now this a general expression for (222222222222.....{2 is repeated n times})^2.
For the given problem,we can find that n=1,2,3,4,5,6,....48.
In the [(eqn 2)],since 1st two terms are already divisible by 9,so we only have
to consider the last term,i.e, 4*(n^2);
Summing up [4*(n^2)] for n=1,2,...48. we get
4{(1^2)+(2^2)+(3^2)+(4^2)+(5^2)+...(48^2)}
=4*[48*(48+1)*(2*48 +1)/6] ; using the formula[1^2 +2^2 +..+n^2=n(n+1)(2n+1)/6]
=4*8*49*97
=4(9- 1)(9*5 +4)(9*11 -2)
=9*B + 4(-1)*(4)*(-2); B an integer, to know its value is not important
=9*B + 32
=9*B + 9*3 + 5;
So, clearly the remainder will be 5.
20. Given that a number Q < 200, calculate sum of all Q such that when Q divided by 5 or 7 givesremainder 2?
1. Find the number of ways in which you can fill a 3x3 grid(with four courners defined as a,b,c,d) if u have 3 white marbles and 6 black marbles.
It is a 3*3 grid, so there are a total of 9 spaces where you can place the marbles . Let us first place the 3 white marbles, that can be done in 9C3 ways. There are 6 spaces available and 6 black marbles are to beplaced in those 6 spaces. So, that can be done in 1 way.
Hence the answer would be 9C3.
X=123456 LCM of 48,98,105 is 11760. So if a number is divisible by 11760, then it will be divisble by 48,98,105. On dividing X by 11760 we get dividend = 10. So for 10 values of y, z is divisible by 11760. Thus for 10 values of y ,z value found is divide by 48,,98,105. These values of Y are 111696, 99936, 88176, 76416, 64656, 52896, 41136, 29376, 17616, 5856
3. There are 6 Bangles each of 4cms in diameter. These are to be placed in a salver(plate), what should be the minimum radius of the salver, so that each bangles are kept without overlapping (banglestouching each other)?
One bangle at the center and the remaining 5 surrounding them, so radius of the center +the diameter of the outer bangle = 4+2=6cms
4. Let Sn denote the sum of first n terms of an A.P. If S2n = 3Sn, then the ratio S3n/Sn is equal to?
We know that Sn = n(n+1)/2
now it is given that S2n = 3Sn
=> 2n(2n+1)/2 = 3n(n+1)/2
=> 2(2n+1) = 3(n+1)
=> 4n+2 = 3n+3
=> n = 1
then S3n/Sn = [3n(3n+1)/2]/[n(n+1)/2] = 3*4/1*2 = 6
now it is given that S2n = 3Sn
=> 2n(2n+1)/2 = 3n(n+1)/2
=> 2(2n+1) = 3(n+1)
=> 4n+2 = 3n+3
=> n = 1
then S3n/Sn = [3n(3n+1)/2]/[n(n+1)/2] = 3*4/1*2 = 6
5. In a strange twist of hearts, P politicians of a country agreed to an average donation of Rs. D each. Q of these politicians, who had pledged an average of Rs. A never donated the pledged money. Which of the following expressions represents the percent of the pledged money that was actually donated.
a)100(QA/PD)
b)100(PD/QA)
c)100-100(QA/PD)
d)100PD-100(QA/PD)
a)100(QA/PD)
b)100(PD/QA)
c)100-100(QA/PD)
d)100PD-100(QA/PD)
We know that Sn = n(n+1)/2
now it is given that S2n = 3Sn
=> 2n(2n+1)/2 = 3n(n+1)/2
=> 2(2n+1) = 3(n+1)
=> 4n+2 = 3n+3
=> n = 1
then S3n/Sn = [3n(3n+1)/2]/[n(n+1)/2] = 3*4/1*2 = 6 pledged money = P*D
Actually donated money = PD-QA
percent of the pledged money thas ws actually donated = 100*(PD-QA)/PD = 100-100(QA/PD)
now it is given that S2n = 3Sn
=> 2n(2n+1)/2 = 3n(n+1)/2
=> 2(2n+1) = 3(n+1)
=> 4n+2 = 3n+3
=> n = 1
then S3n/Sn = [3n(3n+1)/2]/[n(n+1)/2] = 3*4/1*2 = 6 pledged money = P*D
Actually donated money = PD-QA
percent of the pledged money thas ws actually donated = 100*(PD-QA)/PD = 100-100(QA/PD)
6. What is the value of log(e(e(e.....)^1/2)^1/2)^1/2)?
a)0
b)1/3
c)1/2
d)1
b)1/3
c)1/2
d)1
Let x=(e(e(e.........)^1/2)^1/2)^1/2)
now squaring both side we get
x^2=e(e(e(e.........)^1/2)^1/2)^1/2)
i.e., x^2=e*x [since x=(e(e(e.........)^1/2)^1/2)^1/2)]
therefore x=e [since x can not be zero]
Finally, log(x)=log(e)=1
now squaring both side we get
x^2=e(e(e(e.........)^1/2)^1/2)^1/2)
i.e., x^2=e*x [since x=(e(e(e.........)^1/2)^1/2)^1/2)]
therefore x=e [since x can not be zero]
Finally, log(x)=log(e)=1
7. How many values of c in the equation x^2-5x+c result in rational roots which are integers ?
c=4 => x=1,4
c=-6 => x=-1,6
c=-14 => x=-2,7
c=-24 => x=-3,8
c=-36 => x=9,-4
c=-50 => x=-5,10
c=-55 => x=-6,11 and so on...
Hence, Infinite is the answer
c=-6 => x=-1,6
c=-14 => x=-2,7
c=-24 => x=-3,8
c=-36 => x=9,-4
c=-50 => x=-5,10
c=-55 => x=-6,11 and so on...
Hence, Infinite is the answer
8. If 1/a + 1/b + 1/c=1/(a+b+c) where a+b+c != 0, a*b*c != 0 what is the value of (a+b)(b+c)(c+a)?
a)equals 0
b)greater than 0
c)less than 0
d)cannot be determined
a)equals 0
b)greater than 0
c)less than 0
d)cannot be determined
0 (ab+bc+ca)/abc=1/(a+b+c)
=> abc=a2b+ab2+ac2+a2c+b2c+bc2+3abc
=> a2b+ab2+ac2+a2c+b2c+bc2+2abc=0
(a+b)(b+c)(c+a)=(ab+ac+b2+bc)(c+a)
=abc+ac2+b2c+bc2+a2c+a2b+b2a+abc
=0
=> abc=a2b+ab2+ac2+a2c+b2c+bc2+3abc
=> a2b+ab2+ac2+a2c+b2c+bc2+2abc=0
(a+b)(b+c)(c+a)=(ab+ac+b2+bc)(c+a)
=abc+ac2+b2c+bc2+a2c+a2b+b2a+abc
=0
9. PR is a tangent to a circle at point P.Q is another point on the circle such that PQ is the diameter and RQ cuts the circle at point M. If the radius of the circle is 4 units and PR=6 units then find the ratio of the perimeter of triangle PMR to the triangle PQR
a)11/20
b)3/5
c)13/20
d)18/25
a)11/20
b)3/5
c)13/20
d)18/25
Since PR is the tangent,Hence angle QPR is 90.
=>Triangle PQR is rit-angled triangle.
=>hyp QR=10. (Pythagoras Theorem)........(1)
Also PM is perpendicular to QR (Since triangle PQR is made inside circle with diameter as hyp
hence PMQ is rit-angled...itz a property of circle)
=>three triangles PQR,PMQ & PMR are similar to each other.
(For similarity of such type of triangles u can refer any 10th std book).
NOw in that pic consider the triangles PQR & PMR.
PQR~PMR
hence PR/RM=QR/PR.
=>PR^2=RM*QR
putting the values & from(1)
=>RM=18/5....(2)
Also
PQR~PQM
=>PQ/QM=QR/PQ
=>PQ^2=QM*QR
putting the values & from(1)
=>QM=32/5.....(3)
Also,
PQM~PMR
=>PM/QM=RM/PM
=>PM^2=QM*RM
from(2) & (3)...
PM=24/5......(4)
Hence peimeter of triangle PMR=PM+RM+PR
Putting the obtained values perimeter=72/5
Also perimeter of triangle PQR=PQ+QR+PR=24.
Hence th ratio PMR/PQR=72/(5*24)=3/5
=>Triangle PQR is rit-angled triangle.
=>hyp QR=10. (Pythagoras Theorem)........(1)
Also PM is perpendicular to QR (Since triangle PQR is made inside circle with diameter as hyp
hence PMQ is rit-angled...itz a property of circle)
=>three triangles PQR,PMQ & PMR are similar to each other.
(For similarity of such type of triangles u can refer any 10th std book).
NOw in that pic consider the triangles PQR & PMR.
PQR~PMR
hence PR/RM=QR/PR.
=>PR^2=RM*QR
putting the values & from(1)
=>RM=18/5....(2)
Also
PQR~PQM
=>PQ/QM=QR/PQ
=>PQ^2=QM*QR
putting the values & from(1)
=>QM=32/5.....(3)
Also,
PQM~PMR
=>PM/QM=RM/PM
=>PM^2=QM*RM
from(2) & (3)...
PM=24/5......(4)
Hence peimeter of triangle PMR=PM+RM+PR
Putting the obtained values perimeter=72/5
Also perimeter of triangle PQR=PQ+QR+PR=24.
Hence th ratio PMR/PQR=72/(5*24)=3/5
10. The circle O having a diameter of 2cm, has a square inscribed in it.each side of the square is then taken as a diameter to form 4 smaller circles O'.find the total area of all four O' circles which is outside the cirlce O.
a)2
b)pi-2
c)2-pi/4
d)2-pi/2
a)2
b)pi-2
c)2-pi/4
d)2-pi/2
area of cirle O=pi*(1)^2 = pi
Nw the diameter of circle is also the diagonal of the square.
Hence each side of square will be sqrt(2).
=>Area of square=2
since each side of square is also the diameter of other 4 circles.
Hence summation of area of 4 circles=2*pi...........(1)
If u hav drwan its fig u'll find that to obtain the required ans u hav to subtract the area of 4 semi-circles formed on the side of the square from the each of the small portion outside the square.
To get that area of small portion =area of circle O-area of square =pi-2.......(2)
this small portion has to be substracted from the four semi-circles.
Hence, area of 4 semi-circles=2*pi/2= pi......[from (1)]
required ans=total area of 4 semi-circles - area of small portion(from (2))
=pi-(pi-2)
=2.
Nw the diameter of circle is also the diagonal of the square.
Hence each side of square will be sqrt(2).
=>Area of square=2
since each side of square is also the diameter of other 4 circles.
Hence summation of area of 4 circles=2*pi...........(1)
If u hav drwan its fig u'll find that to obtain the required ans u hav to subtract the area of 4 semi-circles formed on the side of the square from the each of the small portion outside the square.
To get that area of small portion =area of circle O-area of square =pi-2.......(2)
this small portion has to be substracted from the four semi-circles.
Hence, area of 4 semi-circles=2*pi/2= pi......[from (1)]
required ans=total area of 4 semi-circles - area of small portion(from (2))
=pi-(pi-2)
=2.
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Important things for elitmus exam preparation students
You can DOWNLOAD e-Litmus question papers and some reference books from the DOWNLOAD section & use them as reference or mock test for your upcoming e-Litmus test
Important things must follow by students who want to crack elitmus
1. Paper level is very tough. elitmus exam is a mock for CAT students.
2. Don’t try to attempt lot of questions.
3. 10 per section in elitmus is ideal attempt to exam.
4. For problem solving section, best scoring part is cryptoarithmetic questions. Try as many question as you have. For crypto arithmetic question don’t refer any books. Just follow this site
If your cryptoarithemetic question are correct then your problem solving section is completed. In DI problem, you can’t write all the questions in time. It is highly impossible.
5. For verbal section, best scoring area is ‘verb form’ type questions. For example download elitmus paper. Now, coming to paragraph questions. Almost 12 questions are there, it is the best scoring area.
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2. arun Sharma is the best book. but leave advance level . solve only average level that will be enough for elitmus.
At last, if you attempt 30 questions in a elitmus exam. If 20 were correct then you can make it. Almost 90 percentile above is in your hand.
Last important thing is sectional percentile is important. So, don’t attempt 20 questions from one section and 10 from another section.
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Here We demostrate a small example...............
Let's understand the way how we will solve these questions with an example:
Let's understand the way how we will solve these questions with an example:
F A C
* H E I
-----------------
E E A G
C J F E
D I J D
------------------
E B F E F G
Step 1:
Determine the characters which can be 0. Although this may not be useful in all cases but keeping the track of it may be useful in many cases.
In our example the multiplicand is FAC and multiplier is HEI
Rule 1: Any leading character will not be 0, so neither of F/H/E/C/D is 0
Rule 2: If any digit among E/I in multiplier is 0 then the corresponding row in multiplication will be all the same digit,
but in our case no row is either EEE/III so neither of E/I is 0
Rule 3: In multiplicand FAC, if C is 0 then the first,second and third row of multiplication will contain the trailing character as C,
but we don't have trailing character in all row as C, so C is no 0
Result of Rule 1, Rule 2 and Rule 3: Characters which can be 0 are - A/B/G/J
Note: Although the rule says that A can be 0 but if you come across situation where you need to consider one among these suspected character as 0,
then consider A(middle character of multiplicand) in the last. Or you can assume it to be non zero for simplicity.
Concluding the step 1: None of the character among multiplier, multiplicand and all leading character in each row will be 0.
So Characters which can be 0 are - B/G/J
Step 2:
Find whether we have any of the following property(in general for all problems):
I * C = I
I * C = C
E * C = E
E * C = C
H * C = H
H * C = C
Remember, they will give problem which has one of these property and this will be starting point of our approach to solve it.
We see that we have the property E * C = E
Now from the rule 3 and 4 we notice that either E is 5 and C is an odd number or E is an even number and C is 6
We will now proceed with the first case when E is 5 and C is an odd number and will consider the second case if we are not able to solve by this assumption.
Also we will have to think differently from here for each problem but practising more and more problem will make it easier.
F A C
* H 5 I
-------------
5 5 A G
C J F 5
D I J D
---------------
5 B F 5 F G
Now look carefully, the last column from left says: D + carry = 5. Now think what could be the maximum carry from the last sum C + I = B ?
Remember the sum of two number will never give a carry more than 1 and sum of three numbers will never give a carry more than 2 except in one case:
U V W
* X Y Z
-------------
9 9 9 9
9 9 9 9
9 9 9 9
-------------
1 1 0 9 8 8 9
Here the 2nd column from left gives a carry equals to 2 when both numbers of the column are same and are equal to 9(not in our case)
So we see that in D + carry = 5, the carry must be 1 and hence D has to be 4
F A C
* H 5 I
-------------
5 5 A G
C J F 5
4 I J 4
-------------
5 B F 5 F G
Also C can be either 3/7/9 and can't be 1 because if C was 1 then trailing character of each row would have been I,5 and H respectively.
Try to minimize the number of values that a character can have. Let's do it for C.
Focus on all C in our problem and see how they are generated. Look at second row of multiplication where:
F A C
* 5
--------
C J F 5
This multiplication give a number which will always be smaller than 5000 so C can be 3 only.
F A 3
* H 5 I
-------------
5 5 A G
3 J F 5
4 I J 4
-------------
5 B F 5 F G
Focus on the second row of multiplication again:
F A 3
* 5
--------
3 J F 5
We can see that F can be either 6 or 7.
Let's proceed with F = 6 and will consider F = 7 if we are stuck and are not able to proceed with F = 6
6 A 3
* H 5 I
-------------
5 5 A G
3 J 6 5
4 I J 4
-------------
5 B 6 5 6 G
Now see the second column from right A+5=6, which implies that A must be 1 and since we haven't gotten any character yet with value 1,
accept the value and proceed further.
6 1 3
* H 5 I
-------------
5 5 1 G
3 J 6 5
4 I J 4
-------------
5 B 6 5 6 G
6 1 3
* 5
--------
3 J 6 5
=> J = 0
6 1 3
* H 5 I
-------------
5 5 1 G
3 0 6 5
4 I 0 4
-------------
5 B 6 5 6 G
6 1 3
* H
--------
4 I 0 4
=> H must be 8 so that when multiplying H with 3 gives unit place digit as 4.
6 1 3
* 8 5 I
-------------
5 5 1 G
3 0 6 5
4 I 0 4
-------------
5 B 6 5 6 G
6 1 3
* 8
--------
4 I 0 4
=> I = 9
6 1 3
* 8 5 9
-------------
5 5 1 G
3 0 6 5
4 9 0 4
-------------
5 B 6 5 6 G
From above we can easily find that G = 7 and B = 2
6 1 3
* 8 5 9
-------------
5 5 1 7
3 0 6 5
4 9 0 4
------------
5 2 6 5 6 7
