greatest 4 digit number which is exactly divisible by 4, 6, 12
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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 factorization :
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
To find the greatest number of 4 digit we have to find the L.C.M of 12, 18, 40,45
Prime factorization :
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
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