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What is the sum of the digits of the largest 4-digit palindromic number which is divisible by 15?
-11
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 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.
What is the largest 4 digit number in base 4?
It is 15.
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 largest 3 digit number that can be divided by 15 25 and 60?
900
What is the largest 3-digit number that can be divided by 15 25 and 60?
900
