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nata0808 [166]
3 years ago
7

When the digits of two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, wha

t is the value of M
Mathematics
1 answer:
hammer [34]3 years ago
6 0

Answer:

The value of M is 91.

Step-by-step explanation:

Suppose T represents the tens digit and U represents the units digit.

Then the numbers M and N are:

M = 10T + U

N = 10U + T

Compute the value of M - N as follows:

M - N = 10T + U - 10U - T = 9T - 9U = 9 (T - U)

Given:

M > N

⇒ T - U > 0

⇒ T > U

Also, M - N is a multiple of 9.

Since M - N is a two-digit number, the possible values of M - N are:

18, 27, 36, 45, 54, 63, and 72

The values 81, 90 and 99 would not be considered since:

9 × 9 = 81

9 × 10 = 90

9 × 11 = 99

T - U = 9 (T - U) = 9 , 10 or 11 respectively and the largest difference for T − U we can get is 9 - 0 = 9 - 0 = 9.

Even this would result in N = 9, which is not a 2-digit number.

It is provided that the integer (M - N) has 12  factors.

Of the possible values of M - N the integer with 12 factors is 72.

The factors of 72 are:

1×72

2×36

3×24

4×18

6×12

8×9

So, M - N = 72.

⇒ M - N = 9 (T - U)

        72 = 9 × 8

⇒ T - U = 8

And since T > U, the values of T and U are:

T = 9 and U = 1

We cannot use T - U = 8 - 0 = 8, because then N will be not a two-digit number.

So, M = 91 and N = 19 and their difference is 72.

Thus, the value of M is 91.

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