Question

Fund the number which when divided by
16, 20,40 leaves the remainder 4
​

Answer 1

$\large\underline{\sf{Solution-}}$

$\sf \: Let \: the \: required \: number \: be \: \bf \: x$

According to statement,

$\sf \: when \:\bf \: x\sf \: is \: divided \: by \: 16, \: it \: leaves \: remaider \: 4$

$\rm :\implies\:(x - 4) \: is \: divisible \: by \: 16. - - - (1)$

Also,

$\sf \: when \:\bf \: x\sf \: is \: divided \: by \: 20, \: it \: leaves \: remaider \: 4$

$\rm :\implies\:(x - 4) \: is \: divisible \: by \: 20. - - - (2)$

Again,

$\sf \: when \:\bf \: x\sf \: is \: divided \: by \: 40, \: it \: leaves \: remaider \: 4$

$\rm :\implies\:(x - 4) \: is \: divisible \: by \: 40. - - - (3)$

• From equation (1), (2), and (3), we concluded that,

$\rm :\implies\:(x - 4) \: is \: divisible \: by \: 16, \: 20 \: and \: 40$

Hence,

$\bf :\implies\:(x - 4) \: = \: LCM \: \: (16, \: 20, \: 40)$

Now,

• Prime factorization of

$\: \: \: \: \: \: \: \bull \: \sf \: 16 = 2 \times 2 \times 2 \times 2 = {2}^{4}$

$\: \: \: \: \: \: \: \bull \: \sf \: 20 = 2 \times 2 \times 5 = {2}^{2} \times 5$

$\: \: \: \: \: \: \: \bull \: \sf \: 40 = 2 \times 2 \times 2 \times 5 = {2}^{3} \times 5$

Hence,

$\bf\implies \:LCM (16, \: 20, \: 40) = {2}^{4} \times 5 = 80$

$\bf\implies \:x - 4 = 80$

$\bf\implies \:x = 84$

Hence,

• 84 is the required number which when divided by 16, 20 and 40, leaves the same remainder 4.

Additional Information :-

Properties of LCM. (least common multiple).

• 1. The L.C.M. of two or more numbers cannot be less than any one of them. For example, the L.C.M. of 2, 3 and 4 is 12 which is not less than any given numbers.

• 2. If a number is the factor of another number, their L.C.M. is the greater number itself. For example, the L.C.M. of 2 and 4 is the greater number 4 itself.

• 3. LCM of two consecutive numbers is always their product. For example, LCM of 2 and 3 is 6.