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how many 4 digit numbers n have the property that the 3 digit number obtained by removing the leftmost digit is one ninth of n

Sagot :

sqdancefan

Answer:

  7

Step-by-step explanation:

We want to find the number 4-digit of positive integers n such that removing the thousands digit divides the number by 9.

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Let the thousands digit be 'd'. Then we want to find the integer solutions to ...

  n -1000d = n/9

  n -n/9 = 1000d . . . . . . add 1000d -n/9

  8n = 9000d . . . . . . . . multiply by 9

  n = 1125d . . . . . . . . . divide by 8

The values of d that will give a suitable 4-digit value of n are 1 through 7.

When d=8, n is 9000. Removing the 9 gives 0, not 1000.

When d=9, n is 10125, not a 4-digit number.

There are 7 4-digit numbers such that removing the thousands digit gives 1/9 of the number.

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