Answer
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Hint:-For solving this question, we must know about H.C.F.
H.C.F.: H.C.F. is the ‘Highest Common Factor’. It means the greatest number which can divide the given numbers.
It is also known as G.C.D.
G.C.D. stands for the Greatest Common Factor.
Complete step-by-step answer:
Let us solve the question now.
So, to solve this question, we need to find a number that divides 120, 165 and 210 exactly leaving remainders 5, 4 and 3 respectively. Which means that we need to find a number that exactly divides (120 – 5), (165 – 4) and (210 – 3).
That means we need to calculate the H.C.F. of (120 – 5), (165 – 4) and (210 – 3) so that these numbers leave remainders 5, 4 and 3.
H.C.F. of 115, 161 and 207
\[\begin{array}{*{35}{l}}
115\text{ }=\text{ }5\times 23 \\
161\text{ }=\text{ }7\times 23 \\
207\text{ }=\text{ }3\times 3\times 23 \\
\end{array}\]
As we can see that the 23 is the only number that is common in the factorization of the three numbers.
Therefore, the H.C.F. of 115, 161 and 207 = 23
So, the required number = 23
Therefore, the greatest number which divides 120, 165 and 210 leaving remainders 5, 4, and 3 respectively is 23.
Hence, the answer of this question is (c) 23.
Note:-Let us now learn about L.C.M.
L.C.M.: L.C.M. stands for the least common multiple, the lowest common multiple. It is also called the smallest common multiple. L.C.M. of two integers ‘a’ and ‘b’ is the smallest positive integer that is divisible by both ‘a’ and ‘b’. This means that ‘a’ and ‘b’ are the factors of that number.
Let us now know about some facts about H.C.F. that can make the calculations easier.
The H.C.F of two or more numbers is smaller than or equal to the smallest number of given numbers.
If the H.C.F of two the numbers ‘a’ and ‘b’ is ‘x’, then the numbers (a + b) and (a -b) are also divisible by ‘x’.
Product of two numbers \[=H.C.F\times L.C.M\text{ }of\text{ }the\text{ }two\text{ }numbers\] .
H.C.F.: H.C.F. is the ‘Highest Common Factor’. It means the greatest number which can divide the given numbers.
It is also known as G.C.D.
G.C.D. stands for the Greatest Common Factor.
Complete step-by-step answer:
Let us solve the question now.
So, to solve this question, we need to find a number that divides 120, 165 and 210 exactly leaving remainders 5, 4 and 3 respectively. Which means that we need to find a number that exactly divides (120 – 5), (165 – 4) and (210 – 3).
That means we need to calculate the H.C.F. of (120 – 5), (165 – 4) and (210 – 3) so that these numbers leave remainders 5, 4 and 3.
H.C.F. of 115, 161 and 207
\[\begin{array}{*{35}{l}}
115\text{ }=\text{ }5\times 23 \\
161\text{ }=\text{ }7\times 23 \\
207\text{ }=\text{ }3\times 3\times 23 \\
\end{array}\]
As we can see that the 23 is the only number that is common in the factorization of the three numbers.
Therefore, the H.C.F. of 115, 161 and 207 = 23
So, the required number = 23
Therefore, the greatest number which divides 120, 165 and 210 leaving remainders 5, 4, and 3 respectively is 23.
Hence, the answer of this question is (c) 23.
Note:-Let us now learn about L.C.M.
L.C.M.: L.C.M. stands for the least common multiple, the lowest common multiple. It is also called the smallest common multiple. L.C.M. of two integers ‘a’ and ‘b’ is the smallest positive integer that is divisible by both ‘a’ and ‘b’. This means that ‘a’ and ‘b’ are the factors of that number.
Let us now know about some facts about H.C.F. that can make the calculations easier.
The H.C.F of two or more numbers is smaller than or equal to the smallest number of given numbers.
If the H.C.F of two the numbers ‘a’ and ‘b’ is ‘x’, then the numbers (a + b) and (a -b) are also divisible by ‘x’.
Product of two numbers \[=H.C.F\times L.C.M\text{ }of\text{ }the\text{ }two\text{ }numbers\] .
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