Q: What is a 3 digit number that is divisible by 1?

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Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.

There is no 4-digit number that is divisible by zero.

the number 3 is divisible by 3

The only 3-digit number that is divisible by 624 is 624.

no

1002 is the smallest 4 digit number divisible by 3 and 2.

-1

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.

Yes.

There are 180 3-digit numbers divisible by five.

To be divisible by 6, the number must be divisible by both 2 and 3:To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8};To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3.As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3.If the number is not divisible by 2 or 3 (or both) then the number is not divisible by 6.examples:126Last digit is even so it is divisible by 2 1 + 2 + 6 = 9 which is divisible by 3, so it is divisible by 3â†’ 126 is divisible by both 2 and 3, so it is divisible by 6124Last digit is even so it is divisible by 2 1 + 2 + 4 = 7 which is not divisible by 3, so it is not divisible by 3â†’ 126 is divisible by 2 but not divisible by 3, so it is not divisible by 6123Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 3 = 6 which is divisible by 3, so it is divisible by 3â†’ 123 is divisible by 3 but not divisible by 2, so it is not divisible by 6121Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 1 = 4 which is not divisible by 3, so it is not divisible by 3â†’ 121 is not divisible by either 2 or 3, so it is not divisible by 6

yes 552 is divisible by 6 and the result is 92... to find out if a number is divisible by 6 then that number must be divisible by 2 and 3.. 1.If a number is divisble by three then the last digit must be divisible by 2 (so that the last digit must be 0,2,4,6,8) in this number the last digit is 2 so that it is divisible by 2 2.If a number is divisble by 3 then the sum of its digit must be divisible by 3. In this case the sum of the digits of 552 is 5+5+2= 12 and 12 is divisible by 3.... If this two rules fit in then that number is divisible by 6

36360 is a 5 digit number that is divisible by both 3 and 9.

The first 3 digit natural number is 100: 100 ÷ 4 = 25 → first 3 digit natural number divisible by 4 is 4 × 25 The last 3 digit natural number is 999: 999 ÷ 4 = 249 r 3 → last 3 digit natural number divisible by 4 is 4 × 249 → number of 3 digit natural numbers divisible by 4 is 249 - 25 + 1 = 225.

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.

1234

The largest 4 digit number that is divisable by 1 2 3 and 4 is........9996

There are 300 three digit numbers that are divisible by neither 2 nor 3. There are 999 - 100 + 1 = 900 three digit numbers. 100 ÷ 2 = 50 → first three digit number divisible by 2 is 50 × 2 =100 999 ÷ 2 = 499 r 1 → last three digit number divisible by 2 is 499 × 2 = 998 → there are 499 - 50 + 1 = 450 three digit numbers divisible by 2. 100 ÷ 3 = 33 r 1 → first three digit number divisible by 3 is 34 × 3 = 102 999 ÷ 3 = 333 → last three digit number divisible by 3 is 333 × 3 = 999 → there are 333 - 34 + 1 = 300 three digit numbers divisible by 3. The lowest common multiple of 2 and 3 is 6, so these have been counted in both those divisible by 2 and those divisible by 3. 100 ÷ 6 = 16 r 4 → first three digit number divisible by 6 is 17 × 6 = 102 999 ÷ 6 = 166 r 3 → last three digit number divisible by 6 is 166 × 6 = 996 → there are 166 - 17 + 1 = 150 → there are 450 + 300 - 150 = 600 three digit numbers that are divisible by either 2 or 3 (or both). → there are 900 - 600 = 300 three digit numbers that are divisible by neither 2 nor 3.

The number 100,002 is divisible by 3. The largest 5 digit number is 99996 (6 less).

108 is the smallest 3-digit number divisible by 2,3,4 & 6

261, 264, or 267

111

21 is not divisible by 2 nor 5, but it is divisible by 3. To be divisible by 2 the last digit must be even (one of {0, 2, 4, 6, 8}; the last digit 1 is not one of these, thus 21 is not divisible by 2 To be divisible by 3 sum the digits and if this total is divisible by 3, then so is the original number; the test can be repeated on the sum, so by repeatedly summing the digits of the totals until a single digit remains only if that single digit is 3, 6 or 9 is the original number divisible by 3. 21 → 2 + 1 = 3 which is divisible by 3 so 21 is divisible by 3. To be divisible by 5, the last digit must be 0 or 5. The last digit of 21 is 1 which is not 0 nor 5, thus 21 is not divisible by 5.

I am a 3 digit number divisible by 7 but not 2 the sum of my digits is 4 what number am I

No. Consider 13 or 23.