## Teaser 2984: Shuffle the cards

**From The Sunday Times, 1st December 2019** [link]

In the classroom I have a box containing ten cards each with a different digit on it. I asked a child to choose three cards and to pin them up to make a multiplication sum of a one-figure number times a two-figure number. Then I asked the class to do the calculation.

During the exercise the cards fell on the floor and a child pinned them up again, but in a different order. Luckily the new multiplication sum gave the same answer as the first and I was able to display the answer using three of the remaining cards from my box.

What was the displayed answer?

[teaser2984]

## Jim Randell 5:05 pm

on29 November 2019 Permalink |If the original sum is

A × BC, then none of the digits in the second sum can be in their original positions, so the second sum is one of:B × CAorC × AB.We can use the [[

`SubstitutedExpression()`

]] solver from theenigma.pylibrary to solve this one.Run:[ @repl.it ]Solution:The result of the two multiplications is 130.The two sums are: 2×65 and 5×26, but we don’t know which was the original one.

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## Jim Randell 9:42 pm

on2 December 2019 Permalink |If we consider the two possible pairs of sums:

Rewriting the first with

A = X, B = Y, C = Zand the second withA = Y, B = Z, C = X, we get:Which are the same, so we can solve the puzzle with an even simpler one liner:

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## GeoffR 7:41 pm

on4 December 2019 Permalink |LikeLike

## GeoffR 4:54 pm

on5 December 2019 Permalink |LikeLike