Swap, Swap, Swap
Let A & B be the 2-digit numbers, where A+B = 120.
If we swap the digits in A (call it A'), A' = A - B.
If we swap the digits in B (call it B'), A - B' = the sum of all digits of A & B.
And A' + B' = 3(B + 1).
Find 100A+B.
The answer is 8436.
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1 solution
We can solve this question by just writing down the first two given conditions.
Let,
A=10p+q
B=10r+s
So
A'=10q+p
B'=10s+r
According to the first statement,
A+B=10p+10r+q+s.......(1)
The second statement states that A'=A-B
10q+p=10p-10r+q-s......(2)
Adding (1) & (2)
120+10q+p=10p+10r+q+s+10p-10r+q-s
120+10q+p=20p+2q
120=19p-8q
Now use your brain.
LHS is even so RHS should also be even . 8q will be even so 19p should be even that's is p should be even
So p={2,4,6,8}
The only possible value of p will be 8 as for other values of p, 120>19p
Hence,
p=8
q=4
r=3
s=6
A=84
B=36
And hence the answer is, 100A+B
=8436