20 if you divide by 17. 19 if you divide by 16. 18 if you divide by 15, 17 if you divide by 14. And so on. In fact any number from 10 to 99. That is, every two digit number.
The largest 2-digit number is 99. When 99 is divided by 4, the quotient is 24 with a remainder of 3. This is because 99 divided by 4 equals 24 with a remainder of 3.
When you divide a 4-digit number by 9, the remainder can only be an integer from 0 to 8. If your answer has a remainder of 2 over 3, it suggests a misunderstanding, as remainders are whole numbers. If you're looking for a 4-digit number that leaves a remainder of 2 when divided by 9, it would be expressed as ( n \equiv 2 \mod 9 ).
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It is 997.
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You can't have a remainder of 6 when you divide by 2! JHC!
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The largest 2-digit number is 99. When 99 is divided by 4, the quotient is 24 with a remainder of 3. This is because 99 divided by 4 equals 24 with a remainder of 3.
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Oh, dude, yeah, totally! A remainder can definitely be a 2-digit number. It's just whatever is left over after you divide one number by another. So, like, if you divide 100 by 3, you get a remainder of 1, which is a 1-digit number. But if you divide 100 by 7, you get a remainder of 2 digits, which is totally cool too.
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It is 997.
mathematics
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Regardless of the dividend (the number being divided), no divisor can produce a remainder equal to, or greater than, itself..... dividing by 4 cannot result in a remainder of 5, for example, Therefore the only single-digit number which can return a remainder of 8 is 9. 35 ÷ 9 = 3 and remainder 8
To find a three-digit number that when divided by 7 gives a remainder of 3 and when divided by 8 also gives a remainder of 3, we can express this mathematically. The number can be written as ( n = 7k + 3 ) for some integer ( k ), and it also needs to satisfy ( n \equiv 3 \mod 8 ). Testing three-digit numbers that meet these conditions, we find that 107 is one such number, as it satisfies both ( 107 \div 7 = 15) remainder 2 and ( 107 \div 8 = 13) remainder 3.
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