What is the smallest number which when decreased by 5 is divisible by 21, 27, 33, and 55?
961314
easy the number is found as such \n\n21 = 3*7\n\n27 = 3^3 \n\n33 = 3 *11\n\n55 = 5 *11\n\nIn order that a number must be divisble by these, it must contain \nall factors. if you look for the samllest avoid repetition \n\n3^3 , 5, 11  =1385\n\nand you add 5 \n\nso the number is 1390
866874
2520,\nwhich is the least common multiple of 2,3,4...,10\n2^3 * 3^2 * 5 * 7
215800
only divisable by 1 and itself\n\nsee prime number pooping bear for examples\n\nhttp://members.surfeu.fi/kklaine/tpnsb/poopbear.html
127181
3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97\n25 is NOT a PRIME NUMBER\n25/5=5
954059
there is a logical way to indicate solutions\n\nfind smallest A such that 1/ A < 3/7\nA= 4\nNow if A = 4\n1/B+1/C = 3/7 -1/4 = 5/28\n\n5/28 = (1+4)/28 = 1/28+1/7 \n            (2+3)/28 no\nif we take A = 5 we may get more solutions\n\n1/B+1/C= 3/7-1/5= 8/35\n= (1+7)/35 = 1/35+1/5 allow if duplicates permitted else no\nno other\n\ntake A =6 and we get\n\n1/B+1/C = 3/7-1/6 = 11/42\nno other solution as 11/42 is not sum of reciprocal
840691
35 = 5*7\n\n210 = 2*3*5*7\n\nTo factor something, start dividing by the smallest prime, if possible, then keep on dividing by larger and larger primes till you run out (i.e., till your remainder is itself prime).\n\nPrimes are 2, 3, 5, 7, 11, 13, 17, etc.\n\nSo, 35 is not divisible by 2 or 3, but it is divisible by 5, with a remainder of 7, so the factorization stops there.  35 = 5*7\n\nNow, 210 is divisible by 2, with remainder 105\n105 is divisible by 3, with remainder 35\n35 = 5*7 (from previous problem)\nSo ... 210 = 2*3*5*7\n\nHope that helped!\n\n~ ♥ ~
146728
no, it is divisible by 1, 3, 23, 29, and itself (with the numbers being reduced as far as possible)
66381
The smallest number that would work would be 5:\n\n5 - (-4) = 9\n\nTherefore the answer would be the following set:\n{all integers greater than 4}\n\ncelesbannon: It says at least 9, not greater than 9, so 5 is a possible answer.
1062332
52\n\n52+\n59+\n64+\n68+\n77\n==320\n\n320 = x + (x-9) + (x-13) + (x-18) + (x-25)\n320 = 5x -65\n385 = 5x\n77 = x\n\nx-25= lowest #\n77-25 = 52
784219
The lowest common multiple, or LCM, of 2 integers is the smallest integer that divides boths\n\nTo work out the LCM of any numbers follow these steps:\n\n1)Prime factorise the numbers\n\n2)Then match the prime divisors of the 2 integers by inserting, if necessary, and primes raised to the power of 0\n\n3)The LCM is the product of all the primes, each raised to the largest exponential\n\nIn this case 35 = 5*7 and 12=2^2*3\n\nThe LCM is 2^2*3*5*7=420
817509
The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both.
802735
You only have about 145 three-digit numbers to test.  The even numbers from 448 to 546 have even squares beginning with 2, from 634 to 706 have even squares beginning with 4, from 776 to 836 have even squares beginning with 6, and from 896 to 948 have even squares beginning with 8.\n\nThe smallest one is 478^2 = 228484, and the largest is 932^2 = 868624.  I would expect there to be 6 or 7 others.
286367
Take the last number off, double it, and subtract it from the rest of the number.  If that is divisible by 7, then the original number is.\n\nFor example 2457\n\n2457 is 245 & 7; 7*2 = 14\n245-14=231\n\nNot sure?  do it again\n\n231 is 23 & 1; 1*2 = 2\n23 - 2 = 21; this is divisible by 7, so 2457 is also divisible by 7.
686152
A common multiple of 2,3,4,5,6,9 and 10 is 64800. But that is not the lowest common multiple which would be a preferable answer. The lowest common multiple is the smallest of the common multiples of a given set of numbers,\n2 = 2 x 1\n3 = 3 x 1\n4 = 2 x 2\n5 = 5 x 1\n6 = 2 x 3\n9 = 3 x 3\n10= 2 x 5\ntheir lowest common multiple would be\n2 x 2 x 3 x 3 x 5 = 180\n180 would be the most ideal answer
638209
1, 2, 3, 6, 7, 14, 21, 42
335428
432 \n216 and 2 \n108 and 2\n54 and 2\n27 and 2 \n9 and 3\n3 and 3 \nThe least common is 3
283335
1,313,131,313 is divisible by 13, and many other numbers