What is the smallest number which when divided by 25 40 and 60 leaves remainder 7 in each case * 1 point?

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Question 1 Playing with numbers Exercise 2.10

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Answer:

Prime factorization of

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

54 = 2 × 3 × 3 × 3

So the required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216

The smallest number which is exactly divisible by 24, 36 and 54 is 216

In order to get remainder as 5

Required smallest number = 216 + 5 = 221

Therefore, the smallest number which when divided by 24, 26 and 54 gives a remainder of 5 each time is 221.

Video transcript

"Welcome to lido homework. My name is leonie Rolla and in today's Q&A video. We are going to solve a word problem. So let us repeat the question together. What is the smallest number which when divided by 24 36 and 54 gives a remainder of 5 each time. So as you can see that in the question, we are supposed to find s smallest number and that smallest number should be common right foot. Before 36 and 54 in what way says in such a way that whenever we divide that number by 24 we divide that number by 36 and we divide that number by 50 for we should get a remainder of 5 each time. So let us think about it, right? How can we do this? So before I start finding that number for who's the remainder after dividing by 24 36 54 we get the remainder as 5 let us And a number that is divisible by 24 by 36 by 54 right? In fact, let us find the least common number that is or least common multiple of 24 36 and 54. So right now what I'm going to do is I'm going to find the LCM of 24 36 and 54. Let us see how we can do that. So 24 the prime factorization of 24 is as follows 212 size twenty four to six Squeals to these ice 6 similarly for 36 it is going to be to 80s is thirty six to nine size 1833 size 9 similarly for 54 2 to the 4 to 7 of 14 three nines ice 27 3 3 0 is 9 therefore 24 can be written as 2 into 2 into 2 into 3. Similarly 36 can be written as 2 into 2 2 into 3 into 3 and few to 4 can be written as 2 into 3 into 3 into 3, right? Let us first find what are the comments here? So we have two two two common and we have three three three common and it has we have three and three common in the second number and the first and second number. We have two common. Therefore. The LCM e is 2/3 This to this tree and the leftovers in the first case that is two in the second case the third case that is three, right? So let us try to multiply this and see how much we will get to these I-66 to size 12 12 into 3 is 36 their son into here, right? So till here it is 36 and 2 into 3 is 6 now, let us see how we can find the value of 36 into 6. So 6 is 6 3 9 6 3 is 18 216 there for two and six is the least common multiple for 24 36 and 54. But the question is not asking us to find the least common multiple. In fact, the question is asking us to find the smallest number the least number which when divided by 24 36 and 54 which will give me five. Now if I divide 2 and 6 by 24 I'm going to get the remainder is 0 similarly if I divide 2 and 6 by 36 I'm going to get the remainder is 0 and also if I divide 216 by 51, I'm going to get the remainder as 0 right. So now what am I supposed to do so that I'll get the remainder as five. My load sticks tells me that I just need to add 5 to 216 which will give me 2 to 1 therefore if I divide 2 to 1 by 24, I'll get the remainder as five. Strangest chicken legs two to one. Let us try to divide it by 24 so we can know that 24 how many times will give me 2 to 1 any idea there is none, right? So let us find just next to that let us find the multiple of 20 for this next 2 to 21 and it will give me it might be to the 6 from here. Also. I can find it very easily to the for photos of eight a tree size 24. They did you guess so if I divide. Let's say Let's Take by 9:00. Okay, 9:00 for size 36 399 to jst in 216 Creek. So if I multiply 24 into 9, I will get it as 216 and I'm getting the remainder is 5 isn't it similarly when I divide 2 to 1 by 36 and when I divide 54 by 3054 221 by 54R still get the remainder is 5 and that is what the question is asking us. Hence now. Now we can conclude that. Hence. 221 is the required smallest number, right? It is the required smallest number. That's it. So what is the Mercury since this is a word problem. You have to give the answer in terms of statements as well. Okay you to give the answer in terms of statement only. Okay guys, that's it for today's Q&A video. If there is any doubt, please do comment below and if you liked the video and for such upcoming videos to subscribe to Lido "

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Answer

Hint: First of all let the least number be N. Then using the division theorem, N = dq + r, write, N = 25a + 9, N = 40b + 9 and N = 60c + 9. Find N by taking LCM of 25, 40 and 9 and adding 9 to it.“Complete step-by-step answer:” Here, we have to find the least number which when divided by 25, 40 and 60 leaves 9 as the remainder in each case. Before solving this question, we must know what division theorem is. Division theorem states that “If ‘n’ is any integer and ‘d’ is a positive integer, there exist unique integers ‘q’ and ‘r’ such that, $n=dq+r$ where 0${\leq}$r<$d$Here, ‘n’ is the number or the dividend, ‘d’ is the divisor, ‘q’ is the quotient and ‘r’ is the remainder.For example, if we divide a number or dividend that is, say 16 by divisor, say 5, we get quotient as 3 and remainder as 1.By division theorem, we can write it as 16 = 5 (2) + 1Now, we have to find the least number which when divided by 25, 40, 60 leaves 9 as the remainder in each case.Here, let us consider the least number to be N. As we know that 25, 40, 60 are divisors and 9 is the remainder in each case. Therefore, by division therefore, we get\[\begin{align}  & N=25a+9....\left( i \right) \\  & N=40b+9....\left( ii \right) \\  & N=60c+9....\left( iii \right) \\ \end{align}\]where a, b, and c are quotients in each case.By subtracting 9 from both sides of the equation (i), (ii) and (iii), we get, \[\begin{align}  & \Rightarrow \left( N-9 \right)=25a \\  & \Rightarrow \left( N-9 \right)=40b \\  & \Rightarrow \left( N-9 \right)=60c \\ \end{align}\]As we know that a, b, and c are integers, therefore we have to find the least value of (N – 9) such that it is a multiple of 25, 40 and 60. That means we have to find the LCM or lowest common multiple of 25, 40 and 60.Now, we will find the LCM of 25, 40 and 60 as follows: