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Q: What are four 3 digit numbers that are divisible by 10?

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No one-digit numbers are divisible by 10. 10 is divisible by 1, 2 and 5.

-10

There are 90,000 such numbers.

There are 5! = 120 such numbers.

The list of numbers that are divisible by 10 is infinite. The first four are: 10, 20, 30, 40 . . .

There are 18 numbers with 2 digits that are divisible by 5. First 2 digit number is 10 → 10 ÷ 5 = 2 → first 2 digit number divisible by 25 is 5 × 2 Last 2 digit number is 99 → 99 ÷ 5 = 19 4/5 → last 2 digit number divisible by 5 is 5 × 19 → There are 19 - 2 + 1 = 18 numbers with 2 digit divisible by 5.

1,234

Any number whose last digit is 0 would be divisible by both these numbers. 1000, 1010, 1020...

There are 9 x 10 x 10 x 10, or 9000 four-digit number that can be formed if the leading digit cannot be zero.

There is no 4-digit number that is divisible by 2356 and 10.

120, 150, 180,210,240,270,300,330,60,390,420,450, 480,510,540,570,600,630,660,690,720,750,780,810,840,870,900,930,860,990

Any multiple of 90 from 1080 to 9990.

All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.

No - but all the numbers divisible by 10 are divisible by 5 !

If the last digit is a zero, the number is divisible by 10.

9000 / 9 = 1000 9000 / 10 = 900

All numbers divisible by 90 are also divisible by 9 and 10

8100

The list of numbers that are divisible by five is infinite. The first four are: 5, 10, 15, 20 . . .

9990

1200 = 3*5*8*10

Multiples of 10 between 1000 and 9990 Any four-digit number ending in zero.

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.

Not evenly. You get 43.2 as the answer. To be divisible by 10, the last digit must be 0 The last digit of 432 is 2 which is not 0, so it is not divisible by 10.

48. Assuming no digit can be used more than once, the two digit numbers divisible by 4 are: 16, 36, 48, 56, 64, 68, 84, 96 8 of them. For any number to be divisible by 4, only the last two digits need be divisible by 4; so for three digit numbers, each of the two digit numbers above can be preceded by any of the remaining 5 digits and still be divisible by 4. → 5 x 8 = 40 three digit numbers are divisible by 4 → 40 + 8 = 48 two or three digit numbers made up of the digits {1, 3, 4, 5, 6, 8, 9} are divisible by 4. If repeats are allowed, there are an extra 2 two digit numbers (44 and 88) and each of the two digit numbers can be preceded by any of the 7 digits, making a total of 7 x 10 + 10 = 80 two and three digits numbers divisible by 4 make up of digits from the given set.