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The sum of 25 consecutive integers is divisible by?
The sum of n consecutive integers is divisible by n when n is odd. It is not divisible by n when n is even. So in this case the answer is it is divisible by 25! Proof: Case I - n is odd: We can substitute 2m+1 (where m is an integer) for n. This lets us produce absolutely any odd integer. Let's look at the sum of any 2m+1 consecutive integers. a + a+d + a+2d + a+3d + ... + a+(n-1)d = n(first+last)/2 (In our problem, the common difference is 1 and this is an arithmetic series.) a + (a+1) + (a+2) + ... + (a+2m) = (2m+1)(2a+2m)/2 = (2m+1)(a+m) It is obvious that this is divisible by 2m+1, our original odd number. That proves case I when n is odd, not for case when it is even. Case II - n is even: We can substitute 2m for n. We have another arithmetic series: a + (a+1) + (a+2) + ... + (a+2m-1) = (2m)(2a+2m-1)/2 = m(2a+2m-1) It is not too hard to prove that this is not divisible by 2m... try it!
What is the measure of the base of a triangle that has side lengths of 3 m and 2 m and a perimeter of 9 m?
The perimeter of a triangle is the sum of its three sides. If two are 3m and 2m then their sum is 3m + 2m = 5m And the remaining side can be calculated by taking this sum from the perimeter → the other side is 9m - 5m = 4m long.
What is 2m x 2m?
It is: 2m times 2m is equivalent to 4m^2
Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
What is greater 2m or 275 cm?
275cm is greater than 2m
