What is the smallest number which when decreased by 5 is divisible by 21, 27, 33, and 55?

Question

I need the solving methods,

Answer 1

easy the number is found as such

21 = 3*7

27 = 3^3

33 = 3 *11

55 = 5 *11

In order that a number must be divisble by these, it must contain
all factors. if you look for the samllest avoid repetition

3^3 , 5, 11 =1385

and you add 5

so the number is 1390

Answer 2

That should be 5 more than the LCM of these numbers. Because LCM is the smallest number divisible by all 4

21 = 3 *7
27 - 3^3
33= 3*11
55= 5* 11
so LCM = 3^3*5*7*11
required number = 3^3*5*7*11+ 5
= 27*5*77 + 5
= 5*(27*77+1)
= 10400

Answer 3

You need to find a number which is a multiple of 21, 27, 33, 55, then add 5.
So it has to be a multiple of 3*7, 3*3*3, 3*11, 5*11. So, looking at each prime, it must be a multiple of 3*3*3, of 5, of 7, and of 11, ie 3*3*3*5*7*11 = 10395. Thus your answer is 10400.

Answer 4

IF YOU MULTIPLY THE 4 NUMBERS YOU HAVE YOU GET:

1029105 HOWEVER THE 21, 27, AND 33 ALL HAVE A FACTOR OF 3 IN COMMON AND 33 AND 55 HAVE A FACTOR OF 11 IN COMMON. THEREFORE TO FIND THE SMALLEST NUMBER YOU MUST DIVIDE BY 3*3 AND 11....I.E. DIVIDE 1029105 BY 99

1029105/99 = 10395

NOW ADD 5 TO GET 10400.

Answer 5

10400

= (5*7*11*27)+5

Answer 6

well..........
21= 3.7
27=3.3.3
33=3.11
55=5.11

considering the factors....... the required num is 5 more than the smallest multiple of 21,27,33,55........
the smallest multiple shud have the non repeating factors....
so the smallest multiple will be....... 27.7.11.5=10395
thus the number is 10395+5= 10400

Answer 7

I think it might be 10400...,just L.C.M the 4 nos & add 5 to it....