bonanova 85 Posted May 2, 2014 Report Share Posted May 2, 2014 Johnny was asked to write the sum of the two fractions p/n and q/n where p, q and n are single-digit numbers. He remembered that the numerators were to be added, but he thought maybe the denominators should be added as well. Also, he decided that putting digits next to each other could suggest addition. So his first try was to say that p/n + q/n = pq/nn. (where pq and nn are two-digit numbers.) Then he remembered the result should not depend on the order of addition, but he saw that p/n + q/n = qp/nn was just as bad. So he decided to use them both. His final representation of the sum was p/n + q/n = (pq + qp)/nn. The puzzle asks: Let p, q and n be chosen at random from {1, 2, 3, 4, 5, 6, 7, 8, 9}. What is the expected error in Johnny's formula? 1 Quote Link to post Share on other sites

0 Solution Rob_G 1 Posted May 2, 2014 Solution Report Share Posted May 2, 2014 There is no error. We know p/n+q/n = (p+q)/n In his formula pq can be written as 10p+q, qp as 10q+p and nn as 10n+n. So (pq+qp)/nn = ([10p+q]+[10q+p])/(10n+n) = (11p+11q)\11n = (11[p+q])/11n = (p+q)/n = p/n+q/n. Quote Link to post Share on other sites

## Question

## bonanova 85

Johnny was asked to write the sum of the two fractions

/pandn/qwheren,pandqare single-digit numbers.nHe remembered that the numerators were to be added, but he thought maybe the denominators should be added as well.

Also, he decided that putting digits next to each other could suggest addition.

So his first try was to say that

/p+n/q=np/q. (wherennpandqare two-digit numbers.)nnThen he remembered the result should not depend on the order of addition,

but he saw that

/p+n/q=nq/pwas just as bad. So he decided to use them both.nnHis final representation of the sum was

/p+n/q= (np+qq)/p.nnThe puzzle asks: Let

,pandqbe chosen at random from {1, 2, 3, 4, 5, 6, 7, 8, 9}.nWhat is the expected error in Johnny's formula?

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