1. **State the problem:** Find the least number $N$ such that: - When divided by 52, remainder is 33. - When divided by 78, remainder is 59. - When divided by 117, remainder is 98. 2. **Rewrite the problem using congruences:** $$N \equiv 33 \pmod{52}$$ $$N \equiv 59 \pmod{78}$$ $$N \equiv 98 \pmod{117}$$ 3. **Convert to a simpler form:** Let $N = x + r$ where $r$ is the remainder. Then: $$N - 33 \equiv 0 \pmod{52}$$ $$N - 59 \equiv 0 \pmod{78}$$ $$N - 98 \equiv 0 \pmod{117}$$ This means: $$N - 33 = 52a$$ $$N - 59 = 78b$$ $$N - 98 = 117c$$ for some integers $a,b,c$. 4. **Express $N$ in terms of $a$:** From the first equation: $$N = 52a + 33$$ Substitute into the second: $$52a + 33 - 59 = 78b \Rightarrow 52a - 26 = 78b$$ $$52a - 78b = 26$$ 5. **Simplify the equation:** Divide entire equation by 2: $$26a - 39b = 13$$ 6. **Solve for integer solutions:** Rewrite: $$26a = 39b + 13$$ Since $26a$ is divisible by 13, right side must be divisible by 13. Check divisibility: $$39b + 13 = 13(3b + 1)$$ So $3b + 1$ must be divisible by 2 (since 26a is even). This is always true for integer $b$. Try to find particular solution: Try $b=1$: $$3(1)+1=4$$ $$26a=13*4=52 \Rightarrow a=2$$ So one solution is $a=2$, $b=1$. 7. **Find $N$ from $a=2$:** $$N=52*2 + 33 = 104 + 33 = 137$$ 8. **Check with third equation:** $$N - 98 = 117c \Rightarrow 137 - 98 = 39 = 117c$$ $$39 = 117c \Rightarrow c = \frac{39}{117} = \frac{1}{3}$$ Not integer, so $N=137$ is not a solution. 9. **General solution for $a$ and $b$:** From step 6, general solution for $a$: $$a = 2 + 3t$$ for integer $t$. Then: $$N = 52a + 33 = 52(2 + 3t) + 33 = 104 + 156t + 33 = 137 + 156t$$ 10. **Use this $N$ in third equation:** $$N - 98 = 117c \Rightarrow 137 + 156t - 98 = 117c$$ $$39 + 156t = 117c$$ Divide entire equation by 3: $$13 + 52t = 39c$$ Rewrite: $$39c - 52t = 13$$ 11. **Solve for integers $t$ and $c$:** Try to find $t$ such that $13 + 52t$ is divisible by 39. Check modulo 39: $$13 + 52t \equiv 0 \pmod{39}$$ Since $52 \equiv 13 \pmod{39}$: $$13 + 13t \equiv 0 \pmod{39}$$ $$13(1 + t) \equiv 0 \pmod{39}$$ Since 13 divides 39, this means: $$1 + t \equiv 0 \pmod{3}$$ $$t \equiv -1 \equiv 2 \pmod{3}$$ 12. **Find smallest $t$ satisfying this:** Smallest $t$ is 2. 13. **Calculate $N$ for $t=2$:** $$N = 137 + 156*2 = 137 + 312 = 449$$ 14. **Verify all conditions:** - $449 \div 52 = 8$ remainder $33$ - $449 \div 78 = 5$ remainder $59$ - $449 \div 117 = 3$ remainder $98$ All conditions satisfied. **Final answer:** $$\boxed{449}$$