Question 521042: EFGHx4=HGFE WHAT NUMBERS ARE E,F,G AND H?
Answer by solver91311(24713)   (Show Source):
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Since both the multiplicand and the product have 4 digits, the 1000s digit of the multiplicand (E) has to be either 1 or 2 otherwise the product would have 5 digits. But if E = 1, then H has to be either 4 or 5, but if H is 4 then E is 6 on the 1s digit end, and if H is 5 then E is 0 on the 1s digit end. Hence, E must be 2. If E is 2, then H is 8 or 9. If H is 9 then E is 6 on the 1s digit end, so H must be 8. That means there was no carry when 4 is multiplied by F. But that means F is either 1 or 2, and since 2 has already been assigned to E, F must be 1. Starting the multiplication process 4 times 8 is 2 carry 3. Since there is a carry of 3 and F is 1, G times 4 must be or end in 8. G cannot be 2 because 2 is already assigned to E, so G must be 7. 4 times 7 is 28 plus 3 is 1 carry 3. Then 4 times 1 plus 3 is 7 as it should be.
E = 2, F = 1, G = 7, and H = 8
And next time you post, don't shout. It is both annoying and rude.
John
My calculator said it, I believe it, that settles it
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