Q.1 find the smallest no. which, when divided by 52, 62 and 72 lraves 2 remainder.
Q.2 what is the greatest 4 digit no. that is exactly divisible by 30 as well as 40?
Answers
Answer:
Q1. 29018
Q2. 9960
Step-by-step explanation:
Steps:
We have to find the smallest number which when divided by 52,62 and 72
that leaves a remainder 2 every time. First, We need to compute,
L.C.M. of 52,62 and 72.
List all prime factors for each number.
Prime Factorization of 52 is: 2 x 2 x 13 => 22 x 131
Prime Factorization of 62 is: 2 x 31 => 21 x 311
Prime Factorization of 72 is: 2 x 2 x 2 x 3 x 3 => 23 x 32
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is 2, 2, 2, 3, 3, 13, 31
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 2 x 3 x 3 x 13 x 31 = 29016
In exponential form:
LCM = 23 x 32 x 131 x 311 = 29016
LCM = 29016
Therefore,
LCM(52, 62, 72) = 29,016
The required smallest number = 29016 + 2 = 29018
So, the smallest number is 29018, which when divided by 52,62 and 72 leaves a remainder of 2 in each case.
Q2.
30=2*3*5
40=2*2*2*5
lcm(30,49)=2*2*2*3*5=120
smallest 5 digit number is 10000
1000/120=83+1/3
83*120=9960
Mark me brainliest :)
Step-by-step explanation:
फाइंड द स्मालेस्ट नंबर दैट कैन बी डिवाइडेड बाय 52 n120