The South African Mathematical Olympiad Third Round

Since the product of the odd integers less than $2014$ contains $25$ as a factor, it is clearly divisible by $25$. Also, since it is odd, its last two digits have to be $25$ or $75$. So the product is either of the form $100n+25$ or of the form $100n+75$. In either case, the last two digits of the squared product (which is the same as the product of the squares) are $25$: $$(100n+25{)}^2=10000n^2+5000n+625=100(100n^2+50n+6)+25$$ or $$(100n+75{)}^2=10000n^2+15000n+5625=100(100n^2+150n+56)+25.$$ In fact, we see that the last three digits have to be $625$.

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Alternative solution.

Our number has $25=5^2$ as a factor, so its last two digits are $00$, $25$, $50$ or $75$. Moreover, it is the square of an odd number, which we can write as $(2n+1{)}^2=4n^2+4n+1$. This means that the remainder upon division by $4$ has to be $1$. Now note that $100n+0$, $100n+25$, $100n+50$ and $100n+75$ leave a remainder of $0$, $1$, $2$, $3$ respectively when divided by $4$. This means that the last two digits are in fact $25$.