Question 521042
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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
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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