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# A little help needed IF you can

## 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

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Full solution:

Given A+B+C, the first two digits are the product of AB. The next two digits are the product of AC. The last two digits are AB+AC-B.

For D+E+F, the last two digits are DE+DF+A.

A=6

B=2

C=5

D=5

E=6

F=7

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A start:

Given A+B+C, the first two digits are the product of AB. The next two digits are the product of AC. Still working on the last two digits.

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Hi Geodood,

logic is :

3+5+7 = 152131

solved as (3*5) (3*7) (15+21-5)

In shorts we can say abc = (a*b)(a*C)(a*b+a*c-b)

Then A+B+C = 123040

In this A = 6 B=2 C=5

Similarly for

IF

9 + 4 + 5 = 364590

2 + 6 + 8 = 121630

3 + 7 + 8 = 212448

here A+B+C = (A*B) (A*C) (A*B+A*C+A)

so

THEN

D + E + F = 303570

D = 5

E = 6

F = 7

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I say thanks to all of you who gave a little hand!

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