Bunuel wrote:
If A,BCD is a four-digit positive integer such that A,BCD is equal to the product of the two-digit number AB and the three digit number ABC, which of the following two-digit numbers cannot be a factor of A,BCD? (Note: the first digit of a number cannot be equal to 0.)
I. AD
II. CD
III. CA
(A) None
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
Kudos for a correct solution.
Ideally, ABC * AB = ABCD should make you think of 1s and 0s. The digits needn't be unique so the first thing that comes to mind is 100*10 = 1000. But since CA and CD are given as two digit numbers, next you should think of 101*10 = 1010 or any such number. This satisfies all conditions so we can analyse using this example.
101 * 10 = 1010
AD = 10 - a factor of 1010
CD = 10 - a factor of 1010
CA = 11 - not a factor of 1010
Now the only thing left to figure is whether it is possible that C1 is a factor of 10C0 for some value of C.
10C0 = 1000 + C0 = 999 + C1
C1 is divisible by C1 but we don't know whether 999 is divisible by C1.
999 = 3^3*37
It has no two digit factor ending in 1 so it will not be divisible by C1.
Answer (C)
In case, 1s and 0s don't come to mind, here is how you could arrive at 10C0:
A bit of visualisation will help:
........A B C
........x A B
__________________
........M N P
.....Q R S x
__________________
.....A B C D
Note the leftmost digits: A*A gives you Q and you get A in the final product.
1*1 = 1 (possible value of Q and A)
2*2 = 4 (even if there were a carry over from previous calculation of M + R, Q+1 = 5, not A(2)). Hence A = 2 is not possible.
3*3 = 9 (even if there were a carry over from previous calculation, Q+1 would be 10 and the final product would be a 5 digit number.)
A cannot be any larger value because then the product would be a 5 digit number. Hence A must be 1 and there must be NO carryover from the previous calculation.
........1 B C
........x 1 B
__________________
........B N P
.....1 B C x
__________________
.....1 B C D
Now note that B+B = B. This is only possible if B = 0 or B = 9 (with a carryover from previous addition). But 9 + 9 + 1 = 19 and that would give a carryover to the next addition which we don't want. So B must be 0
So you get 10C * 10 = 10C0
This is true for all values of C from 1 to 9. (C cannot be 0 because CD is a two digit number)