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Which number is divisible by 2 and 9 that is in between 424 and 449?
Numbers divisible by 9 have the sum of their digits equal to 9 or a multiple of 9. A number divisible by 2 is an even number. If a 3 digit number is 42n then n can only be 3 if the number is divisible by 9 and 423 is not within the specified range. If a 3 digit number is 43n then n must be 2 for it to be divisible by 9.. The number is thus 432 and this is even and so divisible by 2. If the 3 digit number is 44n then n must be 1 and 441 is odd and not divisible by 2. The only valid solution is 432.
Why the divisibility test for 9 is valid and a way to determine whether a three-digit counting number ABC is divisible by 9?
0
How many 10-digit positive integers with distinct digits are multiples of 11111?
-11101
Prove that one of every 3 consecutive positive integers is divisible by 3?
Let three consecutive integers be n, n+1 and n+2. If n is divisible by 3 then n+1 and n+2 cannot be divisible by 3 as these numbers will respectively leave remainders of 1 and 2. If n is not divisible by 3 then it will leave a remainder of 1 or 2. If n leaves a remainder of 1, then n+1 leaves a remainder of 2 and n+2 is therefore divisible by 3. If n leaves a remainder of 2, then n+1 is divisible by 3 and n+2 is not divisible by 3 as it leaves a remainder of 1.
Prove that the two consecutive positive integer is divisible by 2?
There is something missing from your question: what about the two consecutive integers? Is the proof required that one of them is divisible by 2? Or that their product is (which amounts to the same thing)? Showing that one of two consecutive number is divisible by 2: Suppose your two numbers are n and (n+1). If n is divisible by 2, ie n = 2k, the result is shown. Otherwise assume n is not divisible by 2. In this case n = 2m+1. Then: (n+1) = ((2m+1)+1) = 2m + 2 = 2(m+1) which is a multiple of 2 and so divisible by 2. QED. Showing that exactly one of two consecutive integers is divisible by two is shown above with the addition to the first part: "as (n+1) = 2k+1 is not divisible by two and so only n is divisible by 2." To show the product is divisible by 2, show either n is divisible by 2 or (n+1) is as above, then the result follows as one of n and (n+1) is divisible by 2 and so their product is.
What is the number n where if the sum of the digits of the number is divisible by n the number itself is divisible by n?
3 and 9. 93 has a digit sum of 12, initially, which is divisible by 3, but not by 9. So 93 is divisible by 3, but not by 9. 99 has a digit sum of 18, initially, which is divisible by 3 and 9. So 99 is divisible by both 3 and 9.
Which number is divisible by 2 and 9 that is in between 424 and 449?
Numbers divisible by 9 have the sum of their digits equal to 9 or a multiple of 9. A number divisible by 2 is an even number. If a 3 digit number is 42n then n can only be 3 if the number is divisible by 9 and 423 is not within the specified range. If a 3 digit number is 43n then n must be 2 for it to be divisible by 9.. The number is thus 432 and this is even and so divisible by 2. If the 3 digit number is 44n then n must be 1 and 441 is odd and not divisible by 2. The only valid solution is 432.
A number n where if the sum of the digits of the number is divisible by n then the number itself is divisible by n?
n = 1, 3 or 9.
A number n where if the sum of the digits of the number is divisible by n then then the number itself is divisible by n?
This works with 3, and with 9.
What is a number n where if the sum of the digits of the number is divisible by n then the number itself is divisible by n?
That works with 3 and 9.
Why the divisibility test for 9 is valid and a way to determine whether a three-digit counting number ABC is divisible by 9?
0
What is unit digit of 9 to the 20th power?
The units digit of 9n is 9 if n is odd and 1 if n is even. So 1.
What is a number n where if the sum of the digits of the number is divisible by n, then the number itself is divisible by n?
(The assumes that "the number" in the question is not n, although if they are they same number, this is still true.) "If the sum of the digits of the number is divisible by n, then the number itself is divisible by n" is true if n is 3 or if n is 9.
How much two digits divide by 3?
You can find the answer of this question using A.P.(Arithmetic Progression). Its the easiest way. Here's the solution: First 2-digit no. divisible by 3 = 12 Second 2-digit no. divisible by 3 = 15 Last 2-digit no. divisible by 3 = 99 Hence, the 2-digit no.s divisible by 3 form an A.P. with common difference 3 as follows: 12,15,.........................,99 Here, a (First term) = 12 l (Last term) = 99 d (Common difference) = 3 n (No. of terms) = ? Using the formula, l = a + (n-1)d 99 = 12 + (n-1)3 (n-1)3 = 99 - 12 (n-1)3 = 87 n-1 = 87 / 3 n-1 = 29 n = 29 + 1 = 30 Hence, the total no. of 2-digit no.s divisible by 3 are 30.
What is the largest 10 digit number in which the n most digits from the left are divisible by n?
I go for 9,876,545,640 Can't say I am certain, though.
How many 10-digit positive integers with distinct digits are multiples of 11111?
-11101
How many 2 digit numbers are divisible by 5 and 6?
If n is divisible by both 5 and 6, then it should be divisible by 30 (5 * 6). Considering you are asking for only two-digit numbers, the answer(s) would be 30, 60, and 90. So, three numbers.
