find the greatest number that will divide 63,36 and 90 leaving remainder 9 in each case.
Answers
Answered by
0
adding reminder 9 to all the numbers
i) 63 + 9 = 72
ii)36 + 9 = 45
iii)90 + 9 = 99
taking out prime factor s
i) 72 = 2×2×2×3×3
ii) 45 = 5×3×3
iii) 99 = 11×3×3
In this factors the HCF is 3×3 , i.e. 9
: . the required no. is 9
i) 63 + 9 = 72
ii)36 + 9 = 45
iii)90 + 9 = 99
taking out prime factor s
i) 72 = 2×2×2×3×3
ii) 45 = 5×3×3
iii) 99 = 11×3×3
In this factors the HCF is 3×3 , i.e. 9
: . the required no. is 9
Answered by
0
Answer:
27
Step-by-step explanation:
63-9=54
36-9=27
90-9=81
h. C. f. of 54,27,81
27)54(2
54
——
0
27)81(3
81
—
0 so the h. C. f. of 54 ,27,81 is 27
so the greatest number that divides 63,36,90 leaving reminder 9 in each case is 27
Similar questions