Here’s another interesting puzzle for you!

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Thanks for visiting our blog and welcome! Let’s start with the problem statement once again:

Solution:

1. Let’s look at the numbers like – 22, 44, 66 etc. Clearly, if you reverse the order of digits, the numbers remain the same and therefore, their products will remain the same as well.

2. Next, let’s look at the pairs of numbers like – 23 and 32 OR 45 and 54 OR 67 and 76. We can see that the second number in each pair is obtained by reversing the order of digits of the first number. Clearly, for such pairs as well, the product will remain the same.

3. Now, let’s find more of such pairs by doing some algebra!

Let’s assume that the first number is pq and the second number is mn. If we reverse the digits of each, then we obtain qp and nm. Now we want the product of the first pair to be equal to the product of the second pair:

(1op + q)(10m + n) = (10q + p)(10n + m)

100pm + 10pn + 10qm + qn = 100qn + 10qm + 10pn + pm

100pm – pm = 100qn-qn

99pm = 99qn

pm=qn

Thus, if the product of the unit digits is equal to the product of tens digits, then we get pairs of numbers such that if we reverse the order of digits of each, then the product remains the same!

Following are the possible solutions:

12×42 = 21 x24 24×63 = 42×36

12×63 = 21 x36 24×84 = 42×48

13×62 = 31×26 23×96 = 32×69

12×84 = 21×48 26×93 = 62×39

14×82 = 41×28 34×86 = 43×68

13×93 = 31×39 36×84 = 63×48

23×64 = 33×46 46×96 = 64×69

Hope you liked the challenge! At Cuemath, we regularly present challenging and stimulating math puzzles to our students so that their logical thinking and problem-solving abilities are enhanced.

  
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