find the largest number which divided 245,405 and 1029 leaving 5 as remainder in each case
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Answered by
5
We will subtract 5 from 245, 405 and 1029
⛬ 245 - 5 = 240
405 - 5 = 400
1029 - 5 = 1024
Now, we will find the HCF of the obtained numbers i.e. 240, 400 and 1024
240 = 2 × 2 × 2 × 2 × 3 × 5
400 = 2 × 2 × 2 × 2 × 5 × 5
1029 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Common factors = 2 × 2 × 2 × 2
= 16
⛬ The HCF of 240, 400 and 1024 is 16.
⛬ The number is 16
⛬ 245 - 5 = 240
405 - 5 = 400
1029 - 5 = 1024
Now, we will find the HCF of the obtained numbers i.e. 240, 400 and 1024
240 = 2 × 2 × 2 × 2 × 3 × 5
400 = 2 × 2 × 2 × 2 × 5 × 5
1029 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Common factors = 2 × 2 × 2 × 2
= 16
⛬ The HCF of 240, 400 and 1024 is 16.
⛬ The number is 16
Answered by
6
❤❤Here is your answer ✌ ✌
firstly subtract 5 from 245,405 and 1029
245-5=240
405-5=400
1029-5=1025
now factorise
240=2×2×2×2×3×5
405=2×2×2×2×5×5
1029=2×2×2×2×2×2×2×2×2×2
here common factor in( 240,405,1029)=2×2×2×2=16
so ,hence
hcf(240,405,1029)=16
firstly subtract 5 from 245,405 and 1029
245-5=240
405-5=400
1029-5=1025
now factorise
240=2×2×2×2×3×5
405=2×2×2×2×5×5
1029=2×2×2×2×2×2×2×2×2×2
here common factor in( 240,405,1029)=2×2×2×2=16
so ,hence
hcf(240,405,1029)=16
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