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There are (99,999/3) - (10,002/3) + 1 = 33,333 - 3334 + 1 = 30,000 five digit numbers which are evenly divisible by 3. I hope you are not expecting a list.
The first is 10,002. You can find the rest by continually adding 3 until you reach 99,999.
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∙ 12y ago
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How many 3-d How many 3-digit numbers will be there which are divisible by 19igit numbers with atleast one 5 in their digits?
None. 3 digit numbers are not divisible by 19 digit numbers.
How many 3-digit odd palindromes are divisible by 5?
There are 10 3-digit odd palindromes that are divisible by five.
How many 3 digit numbers will be there which are divisible by 19?
The first 3 digit number divisible by 19 is 114 (= 19 x 6) The last 3 digit number divisible by 19 is 988 (= 19 x 52) That means that there are 52 - 6 + 1 = 47 three digit numbers that are divisible by 19.
What is a 4 digit number that is divisible 3 and 9?
All 4 digit numbers that are divisible by 9 are also divisible by 3. The first is 1008 and the last is 9999.
How many 3 digit numbers are divisible by 30?
There are 300.
How many 3 digit number are divisible by 5?
There are 180 3-digit numbers divisible by five.
How many 3 digit numbers are divisible by 6 in all?
There are 151 3-digit numbers that are divisible by 6.
How many 3-d How many 3-digit numbers will be there which are divisible by 19igit numbers with atleast one 5 in their digits?
None. 3 digit numbers are not divisible by 19 digit numbers.
How many 3-digit odd palindromes are divisible by 5?
There are 10 3-digit odd palindromes that are divisible by five.
What five digit number that is divisible by 3?
33333
How many 2 digit numbers are divisible by 3?
Starting at 12 and ending at 99, there are 30 two-digit numbers divisible by three.
How many 3 digit numbers will be there which are divisible by 19?
The first 3 digit number divisible by 19 is 114 (= 19 x 6) The last 3 digit number divisible by 19 is 988 (= 19 x 52) That means that there are 52 - 6 + 1 = 47 three digit numbers that are divisible by 19.
What is a 4 digit number that is divisible 3 and 9?
All 4 digit numbers that are divisible by 9 are also divisible by 3. The first is 1008 and the last is 9999.
How many two - digit numbers are divisible by 3?
There are 30 such numbers.
How many 4 digit numbers are divisible by 3?
2999 of them.
How many 3 digit numbers are divisible by 30?
There are 300.
How many two digit numbers divisible by neither 3 nor 5?
48.The first two digit number is 10, the last two digit number is 99, so there are 99 - 10 + 1 = 90 two digit numbers→ 10 ÷ 3 = 31/3 → first two digit number divisible by 3 is 4 x 3 = 12→ 99 ÷ 3 = 33 → last two digit number divisible by 3 is 33 x 3 = 99→ 33 - 4 + 1 = 30 two digit numbers divisible by 3→ 10 ÷ 5 = 2 → first two digit number divisible by 5 is 2 x 5 = 10→ 99 ÷ 5 = 194/5 → last two digit number divisible by 5 is 19 x 5 = 95→ 19 - 2 + 1 = 18 two digit numbers divisible by 5→ 30 + 18 = 48 two digit numbers divisible by 3 or 5 OR BOTH.The numbers divisible by both are multiples of their lowest common multiple: lcm(3, 5) = 15, and have been counted twice, so need to be subtracted from the total→ 10 ÷ 15 = 010/15 → first two digit number divisible by 15 is 1 x 15 = 15→ 99 ÷ 15 = 69/15 → last two digit number divisible by 15 is 6 x 15 = 90→ 6 - 1 + 1 = 6 two digit numbers divisible by 15 (the lcm of 3 and 5)→ 48 - 6 = 42 two digit numbers divisible by 3 or 5.→ 90 - 42 = 48 two digit numbers divisible by neither 3 nor 5.
