Question

if ones is one more than tenth two-digit number and reversing the number new number is 9 more original number find the number​

Answer 1

Let N be the original number.

Let a be the first (tens) digit of N, and let b be the last (units) of N.

Then N = 10a+b, where a and b are non-negative integers less than 10.

And 10b+a = N/2+1.

So 10b+a = (10a+b)/2+1.

So 10b+a = 10a/2+b/2+1.

So 10b+a = 5a+b/2+1.

So 10b-b/2 = 5a-a+1.

So (10–1/2)b = 4a+1.

So (20/2–1/2)b = 4a+1.

So (20–1)/2×b = 4a+1.

So 19/2×b = 4a+1.

So b = 2/19×(4a+1) = 2×(4a+1)/19 = (2×4a+2×1)/19.

So b = (8a+2)/19.

So 19b = 8a+2.

So 8a = 19b-2.

Since b is an integer, 8a+2 must be an integral multiple of 19, so 8a must be 2 less than an integral multiple of 19. 8a must be even, so 19b must be even. Therefore b must be even (since the parity of the product of two numbers is even if and only if both of the numbers are even). So b is equal to 0, 2, 4, 6 and 8. So we can now try each of these five possible values for B in turn, to see which, if any, give a valid value for a (i.e a non-negative integer less than 10):