Question

A boy saves Rs 4.65 daily than the least number of days in which he will be also to save an exact number of rupees is

Answer 1

Hint: We need to find the least number of days in which he will be also to save an exact number of rupees. We will convert the given rupees to paise and take 100 paise for one rupee and then we take the l.c.m. of 465 and 100 we will get the required number. By dividing the least common multiple by converted paise amount.

Complete step-by-step answer: It is given that the boy saves Rs 4.65 daily.
Rs.4.65= 465paisa
First see the number 4.65.
If we multiply it by any single number\[1,2,3,4, \ldots 8,9\], we will get a number with a decimal point.
Let us take the number 10 then, \[4.65 \times 10 = 46.50\] which is also not an exact number.
The exact number of rupees will be multiple of 100
So, let us find the least common multiple of 465 and 100.
To find the least common multiple of both the numbers we should prime factorize them.
The prime factorization of,
\[100{\rm{ }} = 2 \times 2 \times 5 \times 5\]
\[465 = 3 \times 5 \times 31\]
Least common multiple of 465 and 100 is \[2 \times 2 \times 5 \times 3 \times 5 \times 31 = 9300\]
Thus, the boy saves \[9300paisa = Rs93\]
To find the least number of days to make the saving exact we must divide the total paisa the boy saved by his saving per day.
That is the least number of days required to save 9300 paisa = \[\dfrac{{9300}}{{465}} = 20days\]
Therefore, the least number of days in which he will be also to save an exact number of rupees is 20 days.

Note: LCM: The least common multiple of two integers a and b, usually denoted by
\[lcm\left( {a,\;b} \right)\], is the smallest positive integer that is divisible by both “a” and “b”.
Here we do not use the greatest common divisor to solve the problem because we have to find the least number of days.
Exact number is the whole number which does not have any decimal places.
We have used \[Rs.{\rm{ }}1 = {\rm{ }}100paisa\]