Asked by ithariel whistler age 10
When dividing a 3-digit number by a 1-didgit number,for what divisors can you get a remainder of 8? Explain.
(Taken straight from my worksheet.)
(Taken straight from my worksheet.)
Answers
Answered by
helper
To have a remainder of 8, the divisor must be greater than 8. The only single digit greater than 8 is 9, so 9 must be the divisor.
Since 8 is 1 less than 9, the dividend needs to be one less than a multiple of 9. Candidates are values of 3 digits, as the problem states.
The lowest 3-digit number satisfying the condition is 107 (which is 9x12 - 1); the highest is 998 (which is 9x111 - 1).
Some answers:
107 / 9 = 11 R8
125 / 9 = 13 R8
998 / 9 = 110 R8
from wiki.answers dot com
Since 8 is 1 less than 9, the dividend needs to be one less than a multiple of 9. Candidates are values of 3 digits, as the problem states.
The lowest 3-digit number satisfying the condition is 107 (which is 9x12 - 1); the highest is 998 (which is 9x111 - 1).
Some answers:
107 / 9 = 11 R8
125 / 9 = 13 R8
998 / 9 = 110 R8
from wiki.answers dot com
Answered by
ithariel
thank you
Answered by
helper
you're welcome :)
Answered by
nagelys
8 time 9
Answered by
Shreya
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