$\begingroup$ Show If no repetitions are allowed, how many 9-digit numbers can be formed from the digits 1, 2, 3, …, 9? How many of these are divisible by 4 ? My Attempt: the first part is obviously 9! Now the 2nd part is the tricky one. I saw a similar question where the person who answered sums up all the numbers, in which the sum = 45, but I don't quite understand what to do next. Even if the unit digit is 2,4 or 8, there's no guarantee that it's divisible by 4
asked Oct 1, 2018 at 12:08
 $\endgroup$ $\begingroup$ A number $n$ is divisible by $4$ if and only if $n\mod 100$ is divisible by $4$ (the last two digits). You can find other divisibility rules here: http://www.softschools.com/math/topics/divisibility_rules_2_4_8_5_10/ Now notice that $12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96$ are the only possible two digit combinations that can be made which are divisible by $4$. Thus our answer comes out to be $16\cdot7!$ answered Oct 1, 2018 at 12:19
B2VSiB2VSi 9276 silver badges24 bronze badges $\endgroup$ 1 $\begingroup$ HINT A number is divisible by four when the number made of the last two digits is divisible by four or both equal zero. You just missed the point about two digits by only refering to the unit digit. Therefore you only have to focus on numbers with i.e. $24$ in the end. The rest is irrelevant and can be computed again by using the factorial.
answered Oct 1, 2018 at 12:11
mrtaurhomrtaurho 15.6k14 gold badges29 silver badges68 bronze badges $\endgroup$ 1 How many three-digit numbers are divisible by 9? How many three-digit numbers are divisible by 9? How many three-digit numbers are divisible by 9? Solution: The three-digit numbers divisible by 9 are 108, 117, 126,…., 999. Clearly, these number are in AP. Here. a= 108 and d= 117 — 108 = 9 Let this AP contains n terms. Then. an = 999 =.108+(n – 1) x 9 =999 [an = a + (n -1)d] 9n +99 =999 9n = 999 —99 = 900 n=100 Hence: there are 100 three-digit numbers divisible by 9. Transcribed Image Text:How many 3-digit numbers with digits in {1, ..., 9} (repeats allowed) are divisible by 9? Q: How many numbers between 3000 and 4000, with distinct digits, can be formed using the digits 1,… A: Click to see the answer Q: How many numbers between 100 and 999 (inclusive) are divisible by neither 4 nor 5? 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How many threeThere are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 if no repetition is allowed.
How many 3 number combinations are there in the numbers 0 9?2 Picking a Lock
By comparison, this 3-dial lock (three wheels, each with digits 0-9) has 10 × 10 × 10 = 1, 000 possible combinations. The total number of combinations is not very different, but the Simplex Lock is much harder to pick because it is harder to systematically test each possible combination.
What is the last 3 digit number divisible by 9?We have found that 108 is the smallest 3-digit number that is divisible by 9. Now we will find the largest 3-digit number that is a multiple of 9. The largest 3 digit number is 999, we can see that 999 is divisible by 9 because it satisfies the divisibility rule for 9.
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