Shiva and hari are asked to multiply a three-digit number M with a two-digit number N. Hari multiplied it correctly but Shiva reversed the digits of M as well as N and multiplied the number thus formed. Interestingly both of them got the same answer.
How many such pairs of M and N are possible?
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If both M and N are palindromes than the product of M and N is same as the product of the number formed by taking their digits in the reverse orders. So N can take 9 values (i.e. 11, 22, ....99) Similarly M can take 90 values (i.e. 101, 111, 121, ......, 979, 989, 999) So the combination of M and N can be done in 9 x 90 = 810 ways In addition to these there are other possibilities. M a multiple of 99 can be combined with N, a multiple of 9 to satisfy the given condition. Consider N = 81 and M = 198 ∴ n = 9 x 9 and m = 2 x 99 Reversing M and N NR = 18 = 2 x 9 MR = 891 = 9 x 99 ∴ MN = 9 x 9 x 2 x 99 = 2 x 9 x 9 x 99 Similarly we have (N, M) as (18, 891), (27, 729) (72, 297), (36, 693), (63, 396), (45, 594), (54, 495), (81, 198) So, 8 such combinations are possible. i.e. a total of 810 + 8 = 818 combinations. |
If the digits of N are distinct, M and N are not palindromes So only the latter 8 combinations are possible.
If digits of M are distinct then palindromes combinations are ruled out.
So the maximum value of N is 81 (in the combination of 81 and 198).
There are six digits – 0, 1, 2, 3, 4, and 5. To from 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.
The sum of the numerals 0, 1, 2, 3, 4 and 5 is 15. We know that a 5-digit number is divisible by 3 if and only if the sum of its digits is divisible by '3'. Therefore, we should not use either '0' or '3' while forming the four digit numbers – only then will it be divisible by '3'.
If we do not use '0', then the remaining 5 digits can be arranged in 5! ways = 120 numbers.
If we do not use '3', then the remaining arrangements that are possible without '0' being the first digit are 5! – 4! = 120 – 24 = 96 numbers.
Therefore, there are a total of 120 + 96 = 216 such numbers that exist.