Just Write Them In That Order

Logic Level 1

0 0 = 0 × 0 0 0 = 0 × 0 \LARGE{\begin{aligned} \boxed{\phantom0}\boxed{\phantom0} \; &=& \; \boxed{\phantom0} \times \boxed{\phantom0} \\ \boxed{\phantom0} \boxed{\phantom0} \; &=& \; \boxed{\phantom0} \times \boxed{\phantom0} \\ \end{aligned}}

Is it possible to put one of the integers 1 , 2 , , 8 1, 2, \ldots , 8 into each of the boxes, such that all of these numbers are used and both equations are true?

Clarification : Both the equations above represents a 2-digit integer that can be expressed as the product of two single digit integers.

Yes, it is possible No, it is not possible

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3 solutions

Richard Costen
Jul 16, 2016

Relevant wiki: Modular Arithmetic - Problem Solving - Basic

For interest, one way to arrive at the solution quickly is to notice that the number 5 can only be placed in the leftmost slot. Anywhere else will imply multiplying by 5, which will need another 5 or a 0, which is not allowed. Thus the 2 digit number is between 51 & 58. Only 56 = 7 × 8 56=7\times 8 works. 1 - 4 are remaining and only 3 × 4 3 \times 4 gives a 2 digit number (12). 56 = 7 X 8 12 = 3 X 4 \boxed {56 = 7 X 8} \boxed {12 = 3 X 4}

Yes. Observing that the first digit of the 2-digit integer must be 5 is the most important step. Thank you for your solution

Pi Han Goh - 4 years, 11 months ago

Conveniently, (and uniquely), 12 = 3 × 4 12 = 3 \times 4 and 56 = 7 × 8 56 = 7 \times 8 .

@Brian Charlesworth ahhhhh......interesting! Thank you! (+1)

Noel Lo - 4 years, 11 months ago

Hence, the title of the problem.

Jon Haussmann - 4 years, 7 months ago
Matteo Vassallo
Jul 19, 2016

56=7x8 12=4x3 Used all integers from 1 to 8

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