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What is the sum of the digits of the largest 4-digit palindromic number which is divisible by 15?
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How do you find the greatest number 6 digit exactly divisible by 24 15 and 36?
To find the greatest six-digit number exactly divisible by 24, 15, and 36, first determine the least common multiple (LCM) of these numbers. The LCM of 24, 15, and 36 is 360. The largest six-digit number is 999,999. To find the greatest six-digit number divisible by 360, divide 999,999 by 360 and take the floor of the result, then multiply by 360. This gives you 999,720 as the greatest six-digit number divisible by 24, 15, and 36.
What is a 3 digit number that is divisible by both 3 and 5?
If it's to be divisible by 5, it must end in 5 or 0. And if it's to be divisible by 3, adding the digits together must result in a multiple of 3.The smallest such number is 105; the largest is 990.Since 3 x 5 = 15, all three-digit multiples of 15 would qualify. Note that 105 = 15 x 7, and 990 = 15 x 66.
Which of these cannot be divided evenly into 15 A5 B3 C2 D1 E not of the above?
Divisibility tests:A: 5Last digit of the number is 0 or 5. Last digit of 15 is 5, so 15 is divisible by 15B: 3Sum the digits of the number; if this sum is divisible by 3, so is the original number. 1 + 5 = 6, which is divisible by 3, so 15 is divisible by 3C: 2Number is even (last digit is divisible by 2) 15 is odd, so 15 is not divisible by 2D: 1All numbers are divisible by 1. 15 is a number, so it is divisible by 1.Thus 15 is not divisible by C (2).
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.
How do you know if the number is divisible by 15?
All numbers divisible by 5 (of which 15 is a multiple) have a final digit of 0 or 5. All numbers divisible by 3 (of which 15 is a multiple) have the sum of the digits totalling 3 or a multiple of 3. Therefore, a number is divisible by 15 if the sum of its digits total 3 or a multiple of 3 and its final digit is 0 or 5. Example : 32085 ; 3 + 2 + 0 + 8 + 5 = 18 which is divisible by 3. Final digit 5. This number is divisible by 15. (32085 ÷ 15 = 2139) 7420 : 7 + 4 + 2 + 0 = 13. This number is not divisible by 15.
Is 15 divisible by 5?
To be divisible by 5, the last digit must be 0 or 5.The last digit of 15 is 5, so it is divisible by 5.
What is a 4 digit number divisible by five and nine?
A 4-digit number divisible by both 5 and 9 must be divisible by their least common multiple, which is 45. To find a 4-digit number divisible by 45, we need to find a number that ends in 0 and is divisible by 45. The smallest 4-digit number that fits these criteria is 1005 (45 x 22 = 990, and adding 15 gives us 1005).
What is the divisibility rule for 15?
The number is divisible by both 3 and 5. A number is divisible by 3 if the sum of the digits are and by 5 if the last digit is 0 or 5. Ex: 863,145 Non Ex: 93,460
What is the largest numbers can be divisible by 3 and 5?
There is no such number. If you have any such number, n, that is divisible by 3 and 5 then n + 15 is larger, and is divisible by both. And you can add another 15 to that number, and then to that, for ever more.
What is the largest 4 digit number in base 2?
It is 15.
What is a 5 digit number that is divisible by 9?
A 5-digit number that is divisible by 9 must have its digits sum to a multiple of 9. The smallest 5-digit number that fits this criterion is 10,008 (1 + 0 + 0 + 0 + 8 = 9). Other examples include 10,017, 10,026, and so on, up to the largest 5-digit number that is divisible by 9, which is 99,999.
