Ones place in 10 is 0.
i am a two digit # my tens digit# is 3 times my ones digit #and the sum of my digit is 12 what am i
They are equal to whatever they are equal to.
No, a number with a 9 in the ones place is not always divisible by 3. For a number to be divisible by 3, the sum of its digits must be divisible by 3, regardless of the digit in the ones place. For example, the number 29 has a 9 in the ones place, but its digit sum (2 + 9 = 11) is not divisible by 3.
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All multiples of 10 have zero in the ones place.
There is no four digit number where the ones is twice the tens, the hundreds is five less than the ones, and the thousands is the sum of the tens and hundreds. int ones, tens, hundreds, thousands; for (thousands=1; thousands<10; thousands++) { /**/ for (hundreds=0; hundreds<10; hundreds++) { /**/ /**/ for (tens=0; tens<10; tens++) { /**/ /**/ /**/ for (ones=0; ones<10; ones++) { /**/ /**/ /**/ /**/ if (ones != 2 * tens) break; /**/ /**/ /**/ /**/ if (hundreds != ones - 5) break; /**/ /**/ /**/ /**/ if (thousands != tens + hundreds) break; /**/ /**/ /**/ /**/ printf ("dd\n", thousands, hundreds, tens, ones); /**/ /**/ /**/ } /**/ /**/ } /**/ } }
"The hundreds and the ones are the same digit and their sum is 10" did you say ?Well then, the hundreds and ones digit are both 5.And the tens is 2 less than that, or 3.So the number is . . . . . 535
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1 in the tens place = 10 1 in ones place = 1
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The largest three-digit number with a digit sum of 10 is 910. This can be determined by maximizing the hundreds place first, setting it to 9, and then adjusting the tens and units places to reach a total digit sum of 10. In this case, using 1 for the tens place and 0 for the units place achieves the desired sum.
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