54
54 seems to fit (54 + 5 + 4 = 63).
Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.
100000
In most simple problems like this, I prefer to solve it directly instead of using algebra. If the sum of the digits is 16, the digits can be 7 and 9, 8 and 8, or 9 and 7. So all we have to do is reverse the digits of 79, 88, and 97, to see which reversed number is 18 less than the original. But it should be pretty clear that we don't actually need to do that to all 3, because 97 is the only one that will get smaller by reversing the digits. I'm pretty sure by now that the answer is 97, but to be sure let's reverse the digits and check if the difference is 18: 97-79=18. Yep, 97's the answer! This could also be solved using algebra, like they probably wanted you to do: We'll use T for the tens digit and U for the units digit. Given T+U=16 and 10T+U-18=10U+T, solve. U=16-T 10T+16-T-18=10(16-T)+T 9T-2=160-9T 18T=162 T=9 U=16-9=7 So the original number was 97.
995
54 seems to fit (54 + 5 + 4 = 63).
What is the largest number you can make from the digits 8,6,9,1 explain how you make the number
49 ( + 45 = 94)
Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.
100000
Well, in science you always need significant digits: 0 has no significant digits, so we round to the nearest number with 1 significant digit: namely, -1 or 1.
19
In most simple problems like this, I prefer to solve it directly instead of using algebra. If the sum of the digits is 16, the digits can be 7 and 9, 8 and 8, or 9 and 7. So all we have to do is reverse the digits of 79, 88, and 97, to see which reversed number is 18 less than the original. But it should be pretty clear that we don't actually need to do that to all 3, because 97 is the only one that will get smaller by reversing the digits. I'm pretty sure by now that the answer is 97, but to be sure let's reverse the digits and check if the difference is 18: 97-79=18. Yep, 97's the answer! This could also be solved using algebra, like they probably wanted you to do: We'll use T for the tens digit and U for the units digit. Given T+U=16 and 10T+U-18=10U+T, solve. U=16-T 10T+16-T-18=10(16-T)+T 9T-2=160-9T 18T=162 T=9 U=16-9=7 So the original number was 97.
39
Possibility of two digit no whose sum is 17 89 and 98 Reverse of 89 is 98. 98 is 9 less than the original no 89. 89 is original no
A) If a number has two digits, then the sum of its digits is less than the value of the original two-digit number.
52488