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No... 4 times 3 = 12... That has a 2 in the ones place
Zero. Any five consecutive natural numbers will contain at least one multiple of 2 and at least one multiple of 5, meaning that the product will be a multiple of 10.
No, a multiple of 6 cannot have a ones digit equal to 3. The ones digit of a multiple of 6 will always be even, either 0, 2, 4, 6, or 8, because 6 is divisible by 2.
place value and face value of a number are always equal at ones place.
It is always zero !
No... 4 times 3 = 12... That has a 2 in the ones place
its always the same
All multiples of 10 have zero in the ones place.
Zero. Any five consecutive natural numbers will contain at least one multiple of 2 and at least one multiple of 5, meaning that the product will be a multiple of 10.
No, a multiple of 6 cannot have a ones digit equal to 3. The ones digit of a multiple of 6 will always be even, either 0, 2, 4, 6, or 8, because 6 is divisible by 2.
place value and face value of a number are always equal at ones place.
It is always zero !
The "ones" place is always the last digit in any number so multiples of 5 always have either 5 or 0 (zero) in the "ones" digit position.
yes
It is 0. Two of the first 51 prime numbers are 2 and 5, whose product is 10. When you multiply 10 by any other whole numbers, the final digit (in the ones place value) will be 0.
Eight. It is always the first number to the left side of the decimal point. So if the problem was find the ones place in 520.96, it would be zero
The only number that satisfies both conditions is... 15