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I believe this question is probably more of a level 3 question as the solution is relatively short and elementary (doesn't require advanced knowledge). Firstly, break down the two constants into their prime factorization: $60^{7}$ = $2^{14}$ $\times$ $3^{7}$ $\times$ $5^{7}$ and $25^{14}$ = $5^{28}$ . So just take out the factor $5^{7}$ out of $60^{7}$ to leave $2^{14}$ $\times$ $3^{7}$ , from which we have to calculate the number of divisors. Now if ${n}$ = $p^{a_n}_{n}$ ! = $2^{a_1}$ $\times$ $3^{a_2}$ $\times$ ... $p^{a_n}$ then the number of divisors = [ $a_{1}$ +1][ $a_{2}$ +1]...[ $a_{n}$ +1]. The answer is $\therefore$ 15 $\times$ 8 = $\boxed{120 }$