Question

12. Use Euclid’s division algorithm to find the HCF of 615 and 154. 13. Find the HCF of 867 and 255, using Euclid’s division algorithm. 14. Use Euclid’s division algorithm to find the HCF of 1290 and 228. 15. Use Euclid’s algorithm to find the HCF of 12576 and 4052.

Answer 1

Answer:

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Step-by-step explanation:

12.Let a=615 and b=154

 By Euclids Division Lemma,

     a=bq+r    

   615 = 154*3+153

   154 = 153*1+1

   153=1*153+0

Therfore the HCF of 615 and 154 is 1.

13. let a=867 and b=255

By Euclids Division Lemma,

     a=bq+r

867=255*3+102

255=102*2+51

102=51*2+0

Therfore the HCF of 867 and 255 is 51

14.let a=1290 and b=228

By Euclids Division Lemma,

     a=bq+r

1290 = 228*5 + 150

228 = 150*1 + 78

150 = 78*1 + 72

78 = 72*1 + 6

72 = 6*12 + 0

Therfore the HCF of 1290  and 228 is 6

15.let a=12576 and b=4052

By Euclids Division Lemma,

     a=bq+r

12576=4052*3 +420

4052=420*9+272

420=272*1+148

272=148*1+124

 148=124*1+24

124=24*5+4

 24=4*6+0                           

 Therfore the HCF of  12576 and 4052 is 4                                                                                        

Answer 2

12

13. let a=867 and b=255

13. let a=867 and b=255By Euclids Division Lemma,

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 51

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228By Euclids Division Lemma,

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228By Euclids Division Lemma, a=bq+r

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228By Euclids Division Lemma, a=bq+r1290 = 228*5 + 150

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228By Euclids Division Lemma, a=bq+r1290 = 228*5 + 150228 = 150*1 + 78

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228By Euclids Division Lemma, a=bq+r1290 = 228*5 + 150228 = 150*1 + 78150 = 78*1 + 72

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228By Euclids Division Lemma, a=bq+r1290 = 228*5 + 150228 = 150*1 + 78150 = 78*1 + 7278 = 72*1 + 6

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228By Euclids Division Lemma, a=bq+r1290 = 228*5 + 150228 = 150*1 + 78150 = 78*1 + 7278 = 72*1 + 672 = 6*12 + 0

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228By Euclids Division Lemma, a=bq+r1290 = 228*5 + 150228 = 150*1 + 78150 = 78*1 + 7278 = 72*1 + 672 = 6*12 + 0Therfore the HCF of 1290 and 228 is 6

13. let a=867 and b=255By Euclids Division Lemma, a=bq+r867=255*3+102255=102*2+51102=51*2+0Therfore the HCF of 867 and 255 is 5114.let a=1290 and b=228By Euclids Division Lemma, a=bq+r1290 = 228*5 + 150228 = 150*1 + 78150 = 78*1 + 7278 = 72*1 + 672 = 6*12 + 0Therfore the HCF of 1290 and 228 is 615.let a=12576 and b=4052

a=bq+r

a=bq+r12576=4052*3 +420

a=bq+r12576=4052*3 +4204052=420*9+272

a=bq+r12576=4052*3 +4204052=420*9+272420=272*1+148

a=bq+r12576=4052*3 +4204052=420*9+272420=272*1+148272=148*1+124

a=bq+r12576=4052*3 +4204052=420*9+272420=272*1+148272=148*1+124 148=124*1+24

a=bq+r12576=4052*3 +4204052=420*9+272420=272*1+148272=148*1+124 148=124*1+24124=24*5+4

a=bq+r12576=4052*3 +4204052=420*9+272420=272*1+148272=148*1+124 148=124*1+24124=24*5+4 24=4*6+0

a=bq+r12576=4052*3 +4204052=420*9+272420=272*1+148272=148*1+124 148=124*1+24124=24*5+4 24=4*6+0 Therfore the HCF of 12576 and 4052 is 4