ERRATA for AIMCAT1118 Correction in the choices Directions for question 3 : Answer the question independently of each other. 3. Anoop found the product, P, of two two-digit natural numbers, M and N. He then reversed the digits of each of M and N and found the product of the resultant numbers. Interestingly, he found both products to be the same. If the product of the tens digit of M and the tens digit of N is prime, find the sum of all the possible values of P that Anoop could have obtained. (1) 2604 (2) 2712 (3) 2627 (4) 4684 (5) 4664 Corrected solution 3. Let the two digit numbers be 10a + b and 10c + d ∴ p = (10a + b) (10c + d) = (10b + a) (10d + c) 100ac + 10ad + 10bc + bd = 100bd + 10bc + 10ad + 10ac 99ac = 99bd ac = bd It is known that ac is prime. Whenever the product of two whole numbers is prime, one of them must be 1 and the other must be the prime number. As product of a and c is prime, one of them must be one. The other digit must be prime. ∴The other digit could be 2 or 3 or 5 or 7. ∴ ac = bd = 2 or 3 or 5 or 7. bd must also have one digit as prime and the other digit as 2 or 3 or 5 or 7. The possible products of (ab) and (cd) are (11) (22), (12) (21), (11) (33), (13) (31), (11) (55), (15) (51), (11) (77) and (17) (71) i.e., 242, 252, 363, 403, 605, 765, 847 and 1207. Therefore the sum of all possible products = 4684. Choice (4)
Directions for question 3 : Answer the question independently of
each other.
3. Anoop found the product, P, of two two-digit natural numbers,
M and N. He then reversed the digits of each of M and N and found
the product of the resultant numbers. Interestingly, he found both
products to be the same. If the product of the tens digit of M and
the tens digit of N is prime, find the sum of all the possible
values of P that Anoop could have obtained.
(1) 2604 (2) 2712 (3) 2627 (4) 4684 (5) 4664
Corrected solution
3. Let the two digit numbers be 10a + b and 10c + d p = (10a +
b) (10c + d) = (10b + a) (10d + c) 100ac + 10ad + 10bc + bd = 100bd
+ 10bc + 10ad + 10ac 99ac = 99bd ac = bd It is known that ac is
prime. Whenever the product of two whole numbers is prime, one of
them must be 1 and the other must be the prime number. As product
of a and c is prime, one of them must be one. The other digit must
be prime. The other digit could be 2 or 3 or 5 or 7. ac = bd = 2 or
3 or 5 or 7. bd must also have one digit as prime and the other
digit as 2 or 3 or 5 or 7. The possible products of (ab) and (cd)
are (11) (22), (12) (21), (11) (33), (13) (31), (11) (55), (15)
(51), (11) (77) and (17) (71) i.e., 242, 252, 363, 403, 605, 765,
847 and 1207. Therefore the sum of all possible products = 4684.
Choice (4)
