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svp [43]
2 years ago
6

What number is it that, when multiplied by 18,27,36,45,54,63,72,81, or 99, gives a product in which the first and last figures a

re the same as those in the multiplier, but which when multiplied by 90 gives a product in which the last two figures are the same as those in the multiplier?
Mathematics
1 answer:
allsm [11]2 years ago
6 0

The given multipliers are factors of 9, the possible number will have 1 as first and last digit which gives the number as 101

<h3>How can the correct number be found?</h3>

The given numbers are;

18, 27, 36, 45, 54, 63, 72, 81, 99

Given that the first and the last figures are those of the multipliers, we have;

The first and last and last digit of the number can be 1, which gives;

Possible numbers includes; 11 and 101

By trying the possible numbers, the difference found is presented as follows;

99 × 11 = 1089 (The first and the last digit are not the same as the multiplier)

  • 99 × 101 = 9999 (First and last digit are not the same as the multiplier)

90 × 11 = 990

90 × 101 = 9090

The correct option is therefore 101

  • The number is 101

Learn more about multiplication of numbers here:

brainly.com/question/83239

#SPJ1

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