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Largest no of 5 digits is 99999

99999/99 gives 1010 quotient 9 as remainder

99999-9=99990 is the largest 5 digit no divisible by 99.

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The largest 5-digit number divisible exactly by 99 is 99990.

Q: What largest number of the five digits is divisible by 99?

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99990 ----------------------------------------------------------------------------------------------- Largest 5 digit number is 99999 99999 ÷ 99 = 1010 r 9 → largest 5 digit number divisible by 99 is 99 x 1010 = 99990

98765

99,997

the largest five digit number that can be formed using the digits 7 and 9 is 97,000

97632 is the largest number containing these digits. To make the largest number simply order the numbers in descending order. As it happens this results in an even number so nothing further is necessary.

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99990 ----------------------------------------------------------------------------------------------- Largest 5 digit number is 99999 99999 ÷ 99 = 1010 r 9 → largest 5 digit number divisible by 99 is 99 x 1010 = 99990

98765

123456

99,997

the largest five digit number that can be formed using the digits 7 and 9 is 97,000

97632 is the largest number containing these digits. To make the largest number simply order the numbers in descending order. As it happens this results in an even number so nothing further is necessary.

To be divisible by 15, it must also be divisible by 3 and by 5. To be divisible by 3, the sum of the digits must be divisible by 3; to be divisible by 5, the number must end with a zero or a five. Considering all these criteria, I guess that number would be 1110.

I suggest you try getting different multiples of 41,000, until you find one that has five digits.

1

4,555,558

The only number that ONLY has two and five as factors is 10, which has two digits, so there are no numbers that are three digits long and ONLY have two and five for factors. However, if you are asking what the largest three digit number that has two and five for factors, it would be the largest three digit number with ten as a factor, as ten the the LCM of two and five, and that would be 390.

Because such a number is divisible by ten, and five is a factor of ten. Hence, such a number is also divisible by five.