A remainder is what's left after a division. If I can find a sum that has a one digit answer but a two digit remainder, I've proven it's possible.
915/100= 9 with a remainder of 15. One digit answer, two digit remainder.
So, yes, it's possible.
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Divide the two-digit number by the one-digit number. If the remainder is zero then the 2-digit number is a multiple and if not, it is not.
A 2 digit number divided by a four digit number, such as 2345, will leave the whole 2-digit number as a remainder. It cannot leave a remainder of 1.
You repetitively divide the number by two, taking the remainder as the digit (in binary). When you divide by 2, the remainder will either be 0 or 1.Example: convert 23 (base 10) to binary:23/2 = 11, remainder 1 (this is the ones digit)11/2 = 5, remainder 1 (this is the twos digit)5/2 = 2, remainder 1, (this is the fours digit)2/1 = 1, remainder 0, (this is the eights digit)1/2 = 0, remainder 1, (this is the sixteens digit). So now combine the digits (sixteens is the highest digit in this number):23 (base 10) = 10111 (base 2)
15
63? But it has more than two one-digit factors.