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322222222222222222222222222222×99999999999999999999999999999999322222222222222222222222222222 \times 99999999999999999999999999999999 =

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Q. 322222222222222222222222222222×99999999999999999999999999999999322222222222222222222222222222 \times 99999999999999999999999999999999 =
  1. Set up problem: Set up the multiplication problem.\newlineWe need to multiply two very large numbers: 322222222222222222222222222222322222222222222222222222222222 and 9999999999999999999999999999999999999999999999999999999999999999.
  2. Multiply numbers: Multiply the numbers.\newlineWhen multiplying any number by 9999999999999999999999999999999999999999999999999999999999999999, we can use a shortcut. Multiplying by this number is almost the same as multiplying by 10n10^n (where nn is the number of 99's) and then subtracting the original number. However, since the number of 99's is equal to the number of digits in the first number, we can simplify the process by writing down the first number (322222222222222222222222222222322222222222222222222222222222) and then subtracting it from a number that is the same as the first number but with each digit increased by 11 (i.e., 433333333333333333333333333333433333333333333333333333333333).
  3. Perform subtraction: Perform the subtraction. 433333333333333333333333333333322222222222222222222222222222=111111111111111111111111111111433333333333333333333333333333 - 322222222222222222222222222222 = 111111111111111111111111111111
  4. Verify result: Verify the result.\newlineThe subtraction seems correct, and the pattern of the result (repeating 11's) matches the expected pattern when subtracting a number with repeating 22's from a number with repeating 33's.

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