0
That is correct (unless the number IS 0)
This is because any number that ends in 0 can be divided by 10 (e.g. 10, 20,100, 1864650). Since 10 is a factor of your number, and 2 and 5 are factors of 10, this means that 2, 5 and 10 must all be factors of your number.
There may of course be other factors too like 30 has 3 as a factor also.
Hope this helps
What else can I help you with?
Is 432 divisible by 10?
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.
What is the class digit of any multiple of 10?
The class digit of any multiple of 10 is always 0. This is because multiples of 10 can be expressed in the form (10n), where (n) is an integer, and the last digit of such numbers is always 0. Thus, when considering the class digit, which is the last digit of the number, it consistently remains 0 for all multiples of 10.
Is 195 divisible by 10?
All multiples of 10 end in a zero To be divisible by 10, the last digit must by a 0 The last digit of 195 is a 5 which is NOT a 0, so 195 is NOT divisible by 10.
Is 312 divisible by 10?
All multiples of 10 end in a zero To be divisible by 10, the last digit must by a 0 The last digit of 312 is a 2 which is NOT a 0, so 312 is NOT divisible by 10.
Is 26 divisible by 10?
No. 26 is not evenly divisible by ten. ----------------------------------------------------------------- All multiples of 10 end in a zero To be divisible by 10, the last digit must by a 0 The last digit of 26 is a 6 which is NOT a 0, so 26 is NOT divisible by 10.
Is 432 divisible by 10?
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.
How do you know without multiplying the prime factors that the last digit of the number is 0?
If two of the prime factors are 2 and 5, the last digit of the number will be zero.
What is the class digit of any multiple of 10?
The class digit of any multiple of 10 is always 0. This is because multiples of 10 can be expressed in the form (10n), where (n) is an integer, and the last digit of such numbers is always 0. Thus, when considering the class digit, which is the last digit of the number, it consistently remains 0 for all multiples of 10.
Is 195 divisible by 10?
All multiples of 10 end in a zero To be divisible by 10, the last digit must by a 0 The last digit of 195 is a 5 which is NOT a 0, so 195 is NOT divisible by 10.
Is 312 divisible by 10?
All multiples of 10 end in a zero To be divisible by 10, the last digit must by a 0 The last digit of 312 is a 2 which is NOT a 0, so 312 is NOT divisible by 10.
Is 936 divisible by 10?
All multiples of 10 end in a zero To be divisible by 10, the last digit must by a 0 The last digit of 926 is a 6 which is NOT a 0, so 936 is NOT divisible by 10.
Is 26 divisible by 10?
No. 26 is not evenly divisible by ten. ----------------------------------------------------------------- All multiples of 10 end in a zero To be divisible by 10, the last digit must by a 0 The last digit of 26 is a 6 which is NOT a 0, so 26 is NOT divisible by 10.
Is 117 divisible by 10?
No, 117 is not divisible by 10 because for a number to be divisible by 10, it must end in 0. In this case, 117 does not end in 0, so it is not divisible by 10.
What is the last digit of 21999x32000?
To find the last digit of ( 21999 \times 32000 ), we can focus on the last digits of the numbers involved. The last digit of ( 21999 ) is ( 9 ) and the last digit of ( 32000 ) is ( 0 ). Multiplying these last digits gives ( 9 \times 0 = 0 ). Therefore, the last digit of ( 21999 \times 32000 ) is ( 0 ).
Is 595 divisible by 10?
No. A whole number is (evenly) divisible by 10 only if its last digit is ' 0 '.
Is 32940 divisible by 10?
Yes - since the last digit is 0, we know that 32940 is divisible by 10.
Is 78 divisible by 10?
All multiples of 10 end in a zero To be divisible by 10, the last digit must by a 0 The last digit of 78 is an 8 which is NOT a 0, so 78 is NOT divisible by 10.
