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Is 65, 314, 638, 792 divisible by 24?
\(6+5+3+1+4+6+3+8+7+9+2=54\) and 54 is divisible by 3 because \(5+4=9\). So the number is divisible by 3 . To check for divisibility by 8 , we look at the last three digits, 792 . This is divisible by \(8(792 / 8=99)\). So the number is divisible by both 8 and 3 . So it must be divisible by \(8 * 3=24\).
Is 1, 234, 567, 890, 000 divisible by 750?
Notice that 100 is divisible by 25 , that 1000 is divisible by 125 and that in general \(10^n\) is divisible by \(5^n\). We write any \(k\) digit number as \(m=d_k \ldots d_1=10^n\left(d_k \ldots d_{n+1}\right)+d_n \ldots d_1\). So a number is divisible by \(5^n\) if and only if it’s last \(n\) digits form a number which
is divisible by \(5^n\). Because \(750=2 * 3 * 5^3\), we check for divisibility by 2,3 , and \(5^3\). The last digit is 0 , so the number is divisible by 2 . \(1+2+3+4+5+6+7+8+9=45\) so the number is divisible by 3 . The last three digits are 000 which is divisible by 125 , so the number is divisible by \(5^3\). So, the number is divisible by 750 .
If a number has every one of its digits equal, under what conditions is that number divisible by 11?
Suppose the number has \(n\) digits, all \(k\) ‘s. If \(n\) is even, then \(k-k+k-k+\ldots+k-k=0\) is divisible by 11 , so the number is. If \(n\) is odd, then \(k-k+k-k+\ldots-k+k=k\) which is not divisible by 11. So a number whose digits are all the same is divisible by 11 if and only if it has an even number of digits.
\(
\text { Is } 1001^{10017}-9812521809^2 \text { divisible by } 10 \text { ? }
\)
Any number ending in 1 , when raised to any power, still ends in a 1 . Can you see why? Any number ending in 9 , when squared, also ends in 1 . So the difference of the two numbers above ends in 0 . So it is divisible by 10.
The digits of a number are all 8 ‘s, and it is divisible by 9. What is the least positive integer that fits this description?
The divisibility rule for 9 is that the sum of the digits must be divisible by 9. If the integer consists only of 8’s digits there must be nine of them for the sum to be divisible by 9. 888,888,888 is the smallest integer consisting entirely of 8 ‘s which is divisible by 9.
360 is divisible by both 8 and 9 . How many positive integers less than 360 are also divisible by both 8 and 9?
Integers divisible by both 8 and 9 are divisible by 72. There are 4 multiples of 72 less than 360.
What is the smallest 4-digit number that is divisible by 2, 3, 4, 5, 6, 8, 9,and 10?
The number must end in zero. Let’s assume that to be the smallest it should start with a 1. Since the digits must add up to 9, the last three digits must add up to 8. 1,080 is the smallest four-digit integer divisible by 8 and 9.
Using only 1s and 2s, what is the smallest integer you can create which is divisible by both 3 and 8?
The last digit has to be a 2. Since 12 and 22 don’t work, we need a 3-digit number that is divisible by 8. 112/8 = 14, but it is not divisible by 3. Unfortunately, 122, 212, and 222 are not divisible by 8 so
we must go to a 4-digit number that ends in 112 and whose digits are a multiple of 3. The only 4-digit number that works is 2,112 (we want the sum of the digits to be 6) so it is the smallest.
For the number ABC, each distinct letter represents a different digit. If ABC, CAB, and BCA are all divisible by 6 and 9, find the value of ABC + CAB + BCA.
All three digits must be even for each of ABC, CAB, and BCA to be divisible by 6. The sum of the digits can not be 9 and 27, so we look for three even digits whose sum is 18. The only three distinct digits that work are 4,6, and 8. We can use 468+684+846=1998(Note that 486+648+864=1998)
What digit could be used to fill in the blank to make 452_8 divisible by both 3 and 8 ?
For 452_8 to be divisible by 3, the sum of the digits 4+5+2+_+8 must be divisible by 3, this narrows our choice to 2, 5, and 8. For 452_8 to be divisible by 8, 2_8 must be divisible by 8. Out of 2, 5, and 8, only 8 makes 288 divisible by 8.
How many positive multiples of 3 less than 1,000 use only the digits 2 and/or 4.
There are no 1-digit integers that work. There are two 2-digit integers 24 and 42. 3-digit integers 222 and 444 both work, giving us a total of 4 multiples of 3.
Using only the digits 1,2, and 3 with at least one of each, what is the smallest integer that can be created which is divisible by 8 and 9?
The digit sum must be at least 9 for the integer to be divisible by 9, so we need at least 4 digits: 1, 2, 3, and 3. The last digit must be a 2 for divisibility by 8. We try 1332 and 3132, but neither is divisible by 8. So 3312 is the smallest integer divisible by 8 and 9 which uses each of the digits 1,2,3 at least once.
Using only the digit 2, how many 2 ‘s must be used to create an integer that is divisible by both 9 and 11?
The sum of digits must be divisible by 9, so we try 9 2’s, but there can not be an odd number of 2’s(the sum of alternating sets of digits must be equal) so we must use a minimum of 18 2’s.
What is the sum of all 4-digit multiples of 8 which use each of the digits 1,2,3, and 4 exactly once?
We recognize immediately that the unit digit must be a 2 or a 4. Look at the last 3 digits of a number to determine divisibility by 8. Check only those combinations which end in 2 or 4: 132, 142, 312, 342, 412, 432, 124, 134, 214, 234, 314, 324. From this only 2 gives us multiples of 8: 312 and 432. So the numbers are 4,312 and 1,432. Their sum is 5744.
What is the remainder when 456,564,465,645 is divided by 6?
456,564,465,645 has a digit sum of 4(4+5+6)=60, so it is divisible by 3. It is not even, so it is not divisible by 6. But 3 less than 456,564,465,645 is divisible by 6, so dividing by 6 leaves a remainder of 3.
Using two 5’s and tw0 6’s, it is possible to create four 4-digit numbers that are divisible by 11. What is the sum of these 4 numbers?
The sum of alternating digits must be equal(We can not create a difference of 11 with two 5’s and two 6’s). Four ways this can be done are 5665, 5566, 6556, and 6655. Their sum is 24442.
What digit could fill in the blank to make 89_43 divisible by 11?
Using the divisibility rule for 11, note that for 89×43 to be divisible by 11, 8+x+3 must be equal to 9+4(or the difference of the sums must be divisible by 11, which is not possible in this case). Therefore x =2.
What five-digit multiple of 11 consists entirely of 2 ‘s and 3 ‘s?
The sum of the 2nd and 4th digits must be equal to sum of 1st, 3rd, and 5th digits. So we must use three 2’s and two 3’s (placed in the tens and thousands place) to make 23,232.
What is the largest five-digit multiple of 11?
The largest 5-digit number is 99999. We want the sum of alternating digits to be equal, so we change the unit digit from 9 to 0 to get 99990. This must be the largest 5-digit number divisible by 11 because if we add 11 to it we get a 6-digit number.
Find the smallest positive integer greater than 90,000 that is divisible by 11.
If we use a 2 for the units digit, the sums of alternating digits have a difference of (9+0+2)-(0+0)=11, so 90,002 is divisible by 11.
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