Mẹo What is the smallest number that when divided by 35 56 and 91 leaves remainder of 13 in each case? ?
Kinh Nghiệm về What is the smallest number that when divided by 35 56 and 91 leaves remainder of 13 in each case? Chi Tiết
Hoàng Quang Hưng đang tìm kiếm từ khóa What is the smallest number that when divided by 35 56 and 91 leaves remainder of 13 in each case? được Update vào lúc : 2022-10-03 05:55:25 . Với phương châm chia sẻ Bí quyết Hướng dẫn trong nội dung bài viết một cách Chi Tiết Mới Nhất. Nếu sau khi đọc tài liệu vẫn ko hiểu thì hoàn toàn có thể lại Comments ở cuối bài để Tác giả lý giải và hướng dẫn lại nha.Find the least number which when divides 35, 56 and 91 leaves the same remainder 7 in each case.
Find the least number which when divides 35, 56 and 91 leaves the same remainder 7 in each case.
Answer:
Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91.
Prime factorization of 35, 56 and 91:
35 = 5 × 7
56 = 23 × 7
91 = 7 × 13
LCM = 23 × 5 × 7 × 13
LCM = 3640
Least number which can be divided by 35, 56 and 91 is 3640.
Least number which when divided by 35, 56 and 91 leaves the same remainder,
7 is 3640 + 7 => 3647.
The number is 3647.
First, let’s find the smallest number which is exactly divisible by all 35, 56 and 91.
Which is simply just the LCM of the three numbers.
By prime factorisation, we get
35 = 5 × 7
56 = 23 × 7
91 = 13 × 7
∴ L.C.M (35, 56 and 91) = 23 × 7 × 5 × 13 = 3640
Hence, 3640 is the smallest number which is exactly divisible 28, 42 and 84 i.e. we will get a remainder of 0 in each case. But, we need the smallest number that when divided by 35, 56 and 91 leaves the remainder of 7 in each case.
So that is found by,
3640 + 7 = 3647
∴ 3647 should be the smallest number that when divided by 35, 56 and 91 leaves the remainder of 7 in each case.
What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?
TO FIND: Smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case
L.C.M OF 35, 56 and 91
`35= 5xx7`
`56=2^2xx7`
`91=13xx7`
L.C.M of 35,56 and 91 = `2^2xx5xx7xx13`
=3640
Hence 84 is the least number which exactly divides 28, 42 and 84 i.e. we will get a remainder of 0 in this case. But we need the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case
Therefore
= 3640 +7
= 3640
Hence 3640 is smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case.
Concept: Euclid’s Division Lemma
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hello students today's question is what is the smallest number that when divided by 35 56 and 91 leaves remainder of 7 to find the smallest number which on dividing by 35 56 and 91 leaves remainder of seventh will see that let that number be let let number b capital in of type capital and equal to where is divided by 35 the net will be 35 into a time in plus the remainder which is which it is living right so when the same number is divided by 56 then that type of the number will be 56 and then to be then plus remainder of this is limited by 26 then when that same number is divided by 91.2 C7 or we can write like this
in -7 will be equal to 35 let this be question number 1 and n - 7th equal to 50 60 let this P question number 2 similarity and minus 7 equal to 91c let this P question number 3 now what is this and minus 7 weekend and minus 7 is the LCM of 3 numbers that is 35 5691 56917 and the LCM of 30 and 35 5691 is coming out to be this is 7 into 5 into 18 to 30
cartoon sequel to 4291 which is coming equal to 36 40 so that number is of this type C that 7 was the LCM of n minus 7 and is also the LCM in management consultant equal to 640 now we need to find the smallest number of the smallest smallest number will be that will be what that will be equal to N is equal to 36 40 + 7 that is coming equal to 3647 write this is the smallest number which when divided by 35 56 and 91 leaves a remainder of this is required on thank you
First, let’s find the smallest number which is exactly divisible by all 35, 56, and 91.
Which is simply just the LCM of the three numbers. By prime factorization, we get
35 = 5 × 7
56 = 23 × 7
91 = 13 × 7
∴ L.C.M (35, 56 and 91) = 23 × 7 × 5 × 13 = 3640
Hence, 3640 is the smallest number which is exactly divisible 28, 42 and 84 i.e. we will get a remainder of 0 in each case. But, we need the smallest number that when divided by 35, 56, and 91 leaves the remainder of 7 in each case.
So that is found by,
3640 + 7 = 3647
∴ 3647 should be the smallest number that when divided by 35, 56, and 91 leaves the remainder of 7 in each case.
What is the smallest number that when divided by 35 56 and 91 leaves remainder 13 in each case?
`:. ` Smallest number completely divisibility by 35,36 and 91 is 3640 .
What is the smallest number that when divided by 35 56 and 91 leaves remainder of 7?
The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647. Tải thêm tài liệu liên quan đến nội dung bài viết What is the smallest number that when divided by 35 56 and 91 leaves remainder of 13 in each case?