Let n be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. then sum of the digits in n is: 4 5 6 8
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The equation can be stated as follows:
4665=aN+x4665=aN+x
1305=bN+x1305=bN+x
6905=cN+x6905=cN+x
N is the largest in the above mentioned equation. In order to eliminate X, equations are subtracted from one another
3360=4665−1305=(a−b)N
3360=4665−1305=(a−b)N
5600=6905−1305=(c−b)N
5600=6905−1305=(c−b)N
2240=6905−4665=(c−a)N
2240=6905−4665=(c−a)N
It proves that N is the greatest common factor; therefore one should find similar answer. HCF (2240, 3360, 560) is 1120
4665=aN+x4665=aN+x
1305=bN+x1305=bN+x
6905=cN+x6905=cN+x
N is the largest in the above mentioned equation. In order to eliminate X, equations are subtracted from one another
3360=4665−1305=(a−b)N
3360=4665−1305=(a−b)N
5600=6905−1305=(c−b)N
5600=6905−1305=(c−b)N
2240=6905−4665=(c−a)N
2240=6905−4665=(c−a)N
It proves that N is the greatest common factor; therefore one should find similar answer. HCF (2240, 3360, 560) is 1120
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