Switch, switch, switch

Logic Level 3

Let $A$ and $B$ are 2-digit positive integers, where $A>B$ and $A+B=100$ .

If we switch the digits in $B$ (call it $B'$ ), $B'$ will be the product of both digits of $A$ .

If we switch the digits in $A$ (call it $A'$ ), the absolute difference between the new $A'$ and $B$ will equal to $B' + 1$ .

Find $A \times B$ .

The answer is 1824.

1 solution

Neeraj Kamal
Sep 1, 2015

Let A=10x+y and B=10a+b and A>B and A+B =100 so A must be greater than 100/2 i.e. 50.

Now A.T.Q.

10b+a=xy.

Also, 10y+x-(10a+b)=10b+a+1.

So 10y+x=11(a+b)+1, Now 10x+y is a 2 digit no. So 10y+x cannot be greater than 2 digit. So the possible values of 10y+x should be in the set (01,12,23,34,45,56,67,78,89), now we know that A (10x+y)>50 so the possible values of A should be in the set (54,65,76,87,98) (values are obtained from reversing the values of the possible values of set of 10y+x), now the corresponding values of B are 46,35,24,13,02; the last term i.e. 02 can't take the value of B because it is not a 2 digit no. , now from rest terms the only possible values of A and B should be 76,24. So A×B=76×24=1824

Nice solution

Department 8 - 5 years, 6 months ago

You have to test the case yx < ab as well.

Also from A+B = 100 you can see that (a+b) = (19-x-y) so 10y+x=11(a+b)+1 is equal to 10y+x - 11(19-x-y)+1, or 21y+12x = 210, that mean 7|a and a = 7. The rest follow.

And I don't really understand why there is a condition that ba=x*y

Tran Hieu - 5 years, 5 months ago

Switch, switch, switch

The answer is 1824.

1 solution

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