A number m is such that when it is divided by 30,36 and 45 ,the remainder is always 7.Find the smallest possible value of m

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Step-by-step explanation:

Suppose that you are looking for b, instead of the irrelevant value of m

The statements of this problem suggest us to look for the Least Common Multiple (LCM) of 30, 36 and 45

The number b will be that LCM + 7.

Hence,

30 = 2 x 3 x 5

36 = 2^2 x 3^2

45 = 3^2 x 5

LCM = 180

Therefore the smallest possible value of b (or ‘m’ as per question) is -> LCM+7 that is 180+7= 187

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

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3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !