Answer
454.2k+ views
Hint – In this question first find the greatest 4 digit number, eventually it will be 9999. Now find the L.C.M of the numbers that divide this greatest 4 digit number, that is find the L.C.M of 20, 30, 35 and 40. Then use the concept that the greatest four digit number which is exactly divisible by 20, 30, 35 and 45 is the difference of greatest number and remainder when 9999 is divisible by the L.C.M of the numbers.
Complete Step-by-Step solution:
Let us suppose the numbers w, x, y and z
Now as we know these numbers are exactly divisible by the L.C.M of the numbers w, x, y and z.
Now if we want any remainder after the divisibility say (a) so the required number is (L.C.M of w, x, y and z + remainder (a)).
Now the given numbers is
20, 30, 35 and 45
So first factorize the numbers we have,
So factors of 20 are
$ \Rightarrow 20 = 2 \times 2 \times 5$
Now factors of 30 are
$ \Rightarrow 30 = 2 \times 3 \times 5$
Now factors of 35 are
$ \Rightarrow 35 = 5 \times 7$
And factors of 45 are
$ \Rightarrow 45 = 3 \times 3 \times 5$
Now the L.C.M of given numbers are the product of common factors and remaining factors.
Therefore, L.C.M = \[2 \times 2 \times 3 \times 3 \times 5 \times 7 = 1260\]
Now as we know that the greatest four digit number is 9999.
So the greatest four digit number which is exactly divisible by 20, 30, 35 and 45 is the difference of greatest number and remainder when 9999 is divisible by the L.C.M of the numbers.
So when we divide 9999 by 1260 is
$ \Rightarrow \dfrac{{9999}}{{1260}} = 7\dfrac{{1179}}{{1260}}$
So the remainder is 1179.
So the greatest four digit number which is exactly divisible by 20, 30, 35 and 45 is
$ \Rightarrow 9999 - 1179 = 8820$
Now we want the greatest four digit number which is divisible by 20, 30, 35 and 45 having remainder 12 is
$ \Rightarrow 8820 + 12 = 8832$.
So this is the required answer.
Hence option (C) is correct.
Note – The trick point here was to find the greatest 4 digit number, the leftmost digit can be chosen from numbers (0,1,2,3,4,5,6,7,8,9) , in order to make it highest the leftmost digit must be 9, now in finding this number next digits must be the highest as well as repetition of digits is allowed. In this way 9999 is formed as the highest 4 digit number.
Complete Step-by-Step solution:
Let us suppose the numbers w, x, y and z
Now as we know these numbers are exactly divisible by the L.C.M of the numbers w, x, y and z.
Now if we want any remainder after the divisibility say (a) so the required number is (L.C.M of w, x, y and z + remainder (a)).
Now the given numbers is
20, 30, 35 and 45
So first factorize the numbers we have,
So factors of 20 are
$ \Rightarrow 20 = 2 \times 2 \times 5$
Now factors of 30 are
$ \Rightarrow 30 = 2 \times 3 \times 5$
Now factors of 35 are
$ \Rightarrow 35 = 5 \times 7$
And factors of 45 are
$ \Rightarrow 45 = 3 \times 3 \times 5$
Now the L.C.M of given numbers are the product of common factors and remaining factors.
Therefore, L.C.M = \[2 \times 2 \times 3 \times 3 \times 5 \times 7 = 1260\]
Now as we know that the greatest four digit number is 9999.
So the greatest four digit number which is exactly divisible by 20, 30, 35 and 45 is the difference of greatest number and remainder when 9999 is divisible by the L.C.M of the numbers.
So when we divide 9999 by 1260 is
$ \Rightarrow \dfrac{{9999}}{{1260}} = 7\dfrac{{1179}}{{1260}}$
So the remainder is 1179.
So the greatest four digit number which is exactly divisible by 20, 30, 35 and 45 is
$ \Rightarrow 9999 - 1179 = 8820$
Now we want the greatest four digit number which is divisible by 20, 30, 35 and 45 having remainder 12 is
$ \Rightarrow 8820 + 12 = 8832$.
So this is the required answer.
Hence option (C) is correct.
Note – The trick point here was to find the greatest 4 digit number, the leftmost digit can be chosen from numbers (0,1,2,3,4,5,6,7,8,9) , in order to make it highest the leftmost digit must be 9, now in finding this number next digits must be the highest as well as repetition of digits is allowed. In this way 9999 is formed as the highest 4 digit number.
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