Find the smallest number which when increased by 3 is exactly divisible by 24 36 and 60 ✅ Uy Tín
Mẹo về Find the smallest number which when increased by 3 is exactly divisible by 24 36 and 60 2022
Bùi Thanh Tùng đang tìm kiếm từ khóa Find the smallest number which when increased by 3 is exactly divisible by 24 36 and 60 được Update vào lúc : 2022-12-18 04:40:08 . Với phương châm chia sẻ Mẹo Hướng dẫn trong nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi tham khảo nội dung bài viết vẫn ko hiểu thì hoàn toàn có thể lại Comment ở cuối bài để Mình lý giải và hướng dẫn lại nha.Step 1: Find the LCM Given numbers are 468 and 520Prime factors of 468 are468=2×2×3×3× 13Prime factors of 520 are520=2×2×2×5×13∴LCM of 468 and 520 is LCM468,520=2×2×2×3×3×13×5=4680Step 2: Find the smallest number As per given condition, the number increased by 17 is exactly divisible by 520 and 468∴4680-17=4663Hence, the number is4663.The correct option is C 1439Let us find out the L.C.M. of 12, 18, 24, 32, and 40L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 14401440 = 1439 + 1Hence, 1439 is the smallest number which, when increased by one is exactly divisible by the given numbers.RD Sharma Solutions Class 6 Mathematics Solutions for Playing with numbers Exercise 2.10 in Chapter 2 - Playing with numbersWhat is the smallest number exactly divisible by each of 12 15 20 and 27?What is the smallest number divisible by 10 15 and 20?How will you find the smallest number divisible by 15 and 20?Which is the smallest number that is exactly divisible by 12 15 and 20?What is the smallest number divisible by 10 15 and 20?What is the smallest number which when increased by 17 becomes exactly divisible by 15 20 and 468?What is the smallest number exactly divisible by each of 12 15 20 and 27?What is the smallest number exactly divisible by 15 20 30?SolutionStep 1: Find the LCM Given numbers are 468 and 520Prime factors of 468 are468=2×2×3×3× 13Prime factors of 520 are520=2×2×2×5×13∴LCM of 468 and 520 is LCM468,520=2×2×2×3×3×13×5=4680Step 2: Find the smallest number As per given condition, the number increased by 17 is exactly divisible by 520 and 468∴4680-17=4663Hence, the number is4663.
Nội dung chính Show- Step 1: Find the LCM Given numbers are 468 and 520Prime factors of 468 are468=2×2×3×3× 13Prime factors of 520 are520=2×2×2×5×13∴LCM of 468 and 520 is LCM468,520=2×2×2×3×3×13×5=4680Step 2: Find the smallest number As per given condition, the number increased by 17 is exactly divisible by 520 and 468∴4680-17=4663Hence, the number is4663.The correct option is C 1439Let us find out the L.C.M. of 12, 18, 24, 32, and 40L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 14401440 = 1439 + 1Hence, 1439 is the smallest number which, when increased by one is exactly divisible by the given numbers.RD Sharma Solutions Class 6 Mathematics Solutions for Playing with numbers Exercise 2.10 in Chapter 2 - Playing with numbersWhat is the smallest number exactly divisible by each of 12 15 20 and 27?What is the smallest number divisible by 10 15 and 20?How will you find the smallest number divisible by 15 and 20?Which is the smallest number that is exactly divisible by 12 15 and 20?What is the smallest number divisible by 10 15 and 20?What is the smallest number which when increased by 17 becomes exactly divisible by 15 20 and 468?What is the smallest number exactly divisible by each of 12 15 20 and 27?What is the smallest number exactly divisible by 15 20 30?SolutionStep 1: Find the LCM Given numbers are 468 and 520Prime factors of 468 are468=2×2×3×3× 13Prime factors of 520 are520=2×2×2×5×13∴LCM of 468 and 520 is LCM468,520=2×2×2×3×3×13×5=4680Step 2: Find the smallest number As per given condition, the number increased by 17 is exactly divisible by 520 and 468∴4680-17=4663Hence, the number is4663.The correct option is C 1439Let us find out the L.C.M. of 12, 18, 24, 32, and 40L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 14401440 = 1439 + 1Hence, 1439 is the smallest number which, when increased by one is exactly divisible by the given numbers.RD Sharma Solutions Class 6 Mathematics Solutions for Playing with numbers Exercise 2.10 in Chapter 2 - Playing with numbersWhat is the smallest number that when decreased by 3 is divisible by 24 32 36 and 54?What is the smallest number which when increased by 3 is divisible by?What is the smallest number which when increased by 3 is exactly divisible by 15 18 20 and 24?What is the smallest number which when increased by it is divisible by 21 27 33 and 55?
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Solution
The correct option is C 1439Let us find out the L.C.M. of 12, 18, 24, 32, and 40L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 14401440 = 1439 + 1Hence, 1439 is the smallest number which, when increased by one is exactly divisible by the given numbers.
SolveTextbooksQuestion PapersNội dung chính Show
- RD Sharma Solutions Class 6 Mathematics Solutions for Playing with numbers Exercise 2.10 in Chapter 2 - Playing with numbersWhat is the smallest number exactly divisible by each of 12 15 20 and 27?What is the smallest number divisible by 10 15 and 20?How will you find the smallest number divisible by 15 and 20?Which is the smallest number that is exactly divisible by 12 15 and 20?
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RD Sharma Solutions Class 6 Mathematics Solutions for Playing with numbers Exercise 2.10 in Chapter 2 - Playing with numbers
Question 1 Playing with numbers Exercise 2.10
What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?
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 "
What is the smallest number that when decreased by 3 is divisible by 24 32 36 and 54?
Hence, 221 is required number.What is the smallest number which when increased by 3 is divisible by?
To find : The smallest number ? As number is increased by 3. So, Therefore, the smallest number which when increased by 3 and is divisible by 27, 48, 51 is 7341.What is the smallest number which when increased by 3 is exactly divisible by 15 18 20 and 24?
Loved by our community So the answer is 942.What is the smallest number which when increased by it is divisible by 21 27 33 and 55?
We need to find the LCM of the given numbers, LCM of 21, 27, 33, and 55 = 10395. 10403 when decreased by 8 will be divisible by 21,27,33 and 55. Tải thêm tài liệu liên quan đến nội dung bài viết Find the smallest number which when increased by 3 is exactly divisible by 24 36 and 60