Question

If the sum of three whole numbers taken in pairs are 43, 50 and 63, then the largest five digit number divisible by the three whole numbers arranged in ascending order leaves the remainder 11, 24 and 31 respectively is​

Answer 1

Answer:

2

Step-by-step explanation:

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Answer 2

Given:

If the sum of three whole numbers taken in pairs are 43, 50 and 63

To Find:

then the largest five-digit number divisible by the three whole numbers arranged in ascending order leaves the remainder 11, 24 and 31 respectively is​

Solution:

Let the numbers be a,b,c then we can formulate equations in the form,

$a+b=43$

$b+c=50$

$a+c=63$

Now solving the following equations using the elimination rule we will get the following values,

a=28

b=15

c=35

Now it is given that when these numbers are arranged in ascending order that is 15,28,35 when dividing the largest 5 digit number leaves the remainder as 11,24,31 respectively. Before solving the problem we should know that when divided by 2 or more divisors leaving a different remainder but the common difference is same the then we can denote the number N as,

$N=LCM*k-difference$

in the given case we can that the common difference is 4, so

$N=LCM(15,28,35)*k-4\\=420k-4$

So the largest 5 digit number will be found when k=238, now

$N=420*238-4\\=99960-4\\=99956$

Hence, the largest 5 digit number is 99956.