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## Question

Hi everyone!

Today I came across with this riddle that seems simple, but I haven't got a clue on how to solve.

IF

3+5+7 = 152131

4+8+6 = 322448

6+2+9 = 125464

THEN

A + B + C = 123040

--------------------------

IF

9 + 4 + 5 = 364590

2 + 6 + 8 = 121630

3 + 7 + 8 = 212448

THEN

D + E + F = 303570

Thanks in advance

Edited by Geodood

## Recommended Posts

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The first 2 digits are the product of the first 2 numbers in the sum: 3*5=15. Then you multiply the first and last numbers: 3*7=21. Finally you add 15+21 and subtract the centre number: 15+21=36, 36-5=31. So if A+B+C=123040, then AB=12, AC=30, and B=42-40=2, so A=6, C=5. In the second version, instead of subtracting the 2nd number, you're adding the first: 3*7=21, 3*8=24, 21+24+3=48. So if D+E+F=303570, DE=30, DF=35, and 65+D=70, making D=5, E=6, F=7.

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