find the greatest number of four digits which is exactly divisible by 12,18,40,45
Answers
To find the greatest 4 digit number which is exactly divisible by 12, 18, 40 and 45, first we have to compute the LCM of 12, 18, 40 and 45.
_______________
2 | 12, 18, 40, 45
|_______________
2 | 6, 9, 20, 45
|_______________
2 | 3, 9, 10, 45
|_______________
3 | 3, 9, 5, 45
|_______________
3 | 1, 3, 5, 15
|_______________
5 | 1, 1, 5, 5
|_______________
| 1, 1, 1, 1
The LCM of 12, 18, 40 and 45 = 2*2*2*3*3*5 = 360
Now, we will divide the greatest 4 digit number by 360.
The greatest 4 digit number is 9999.
______
360) 9999 (27
720
_____
2799
2520
______
279 - Remainder
______
Now, we have to subtract the remainder, which is 279, from 9999.
⇒ 9999 - 279 = 9720
So, 9720 is the required greatest 4 digit number which is exactly divisible by 12, 18, 40 and 45.
Answer:
9720
Step-by-step explanation:
We know that the Greatest number of four digits = 9999
To find the greatest number of 4 digit we have to find the L.C.M of 12, 18, 40,45
Prime factorisation :
12= 2×2×3= 2²×3¹
18= 2×3×3= 2¹× 3²
40= 2×2×2×5= 2³×5
45= 3×3×5= 3²×5
L.C.M of 12 ,18,40,&45= 2³×3²×5
L.C.M of 12 ,18,40,&45= 2×2×2×3×3×5
L.C.M of 12 ,18,40,&45= 360
Now divide the largest 4 digit number 9999 by 360
9999 ÷ 360
9999= 360× 27 +279
Remainder= 279
Then Subtract the remainder from 9999
9999- 279= 9720
Hence, greatest number of four digits which is divisible by 12, 18, 40 & 45
= 9720