She has to find four positive integers such that where and are odd numbers and are prime numbers such that a^3+b^3+c^3+d^3=n
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p and q are two positive integers whose l.c.m.
is r2 s 4 t 2. This ?rst of all means that neither p nor q can have any prime
factor besides r, s and t. So each of them is a product of powers of some
of these three primes. We can therefore write p, q in the form
p = ra sb tc and q = ru sv tw (1)
where a, b, c, u, v, w are non-negative integers. Then the l.c.m., say e, of p and q is given by
e = ri sj tk (2)
where
i = max{a, u}, j = max{b, v} and k = max{c, w} (3)
This is the key idea of the problem. The problem is now reduced to ?nding
the number of triplets of ordered pairs of the form {(a, u), (b, v), (c, w)}
where a, b, c, u, v, w are non-negative integers that satisfy
max{a, u} = 2, max{b, v} = 4 and max{c, w} = 2 (4)
Let us see in how many ways the ?rst entry of this triplet, viz., (a, u)
can be formed. We want at least one of a and u to equal 2. If we let a = 2,
then the possible values of u are 0, 1 and 2. These are three possibilities.
Similarly, with u = 2 there will be three possibilities, viz. a = 0, 1 or 2.
So, in all the ?rst ordered pair (a, u) can be formed in 6 ways. But the
possibility (2, 2) has been counted twice. So, the number of ordered pairs
of the type (a, u) that satisfy the ?rst requirement in (4) is 5 and not 6.
By an entirely analogous reasoning, the number of ordered pairs of
the form (b, v) which satisfy the second requirement in (4) is 2 × 5 − 1,
i.e. 9 while that of ordered pairs of the type (c, w) satisfying the third
requirement in (4) is 5. But the ways these three ordered pairs are formed
are completely independent of each other. So the total number of triplets
of ordered pairs of the form {(a, u), (b, v), (c, w)} where a, b, c, u, v, w are
non-negative integers that satisfy (4) is 5 ×9 ×5 = 225.
is r2 s 4 t 2. This ?rst of all means that neither p nor q can have any prime
factor besides r, s and t. So each of them is a product of powers of some
of these three primes. We can therefore write p, q in the form
p = ra sb tc and q = ru sv tw (1)
where a, b, c, u, v, w are non-negative integers. Then the l.c.m., say e, of p and q is given by
e = ri sj tk (2)
where
i = max{a, u}, j = max{b, v} and k = max{c, w} (3)
This is the key idea of the problem. The problem is now reduced to ?nding
the number of triplets of ordered pairs of the form {(a, u), (b, v), (c, w)}
where a, b, c, u, v, w are non-negative integers that satisfy
max{a, u} = 2, max{b, v} = 4 and max{c, w} = 2 (4)
Let us see in how many ways the ?rst entry of this triplet, viz., (a, u)
can be formed. We want at least one of a and u to equal 2. If we let a = 2,
then the possible values of u are 0, 1 and 2. These are three possibilities.
Similarly, with u = 2 there will be three possibilities, viz. a = 0, 1 or 2.
So, in all the ?rst ordered pair (a, u) can be formed in 6 ways. But the
possibility (2, 2) has been counted twice. So, the number of ordered pairs
of the type (a, u) that satisfy the ?rst requirement in (4) is 5 and not 6.
By an entirely analogous reasoning, the number of ordered pairs of
the form (b, v) which satisfy the second requirement in (4) is 2 × 5 − 1,
i.e. 9 while that of ordered pairs of the type (c, w) satisfying the third
requirement in (4) is 5. But the ways these three ordered pairs are formed
are completely independent of each other. So the total number of triplets
of ordered pairs of the form {(a, u), (b, v), (c, w)} where a, b, c, u, v, w are
non-negative integers that satisfy (4) is 5 ×9 ×5 = 225.
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