Eggs in basket

If the eggs in a basket are removed twice at a time, one will remain. Two eggs will remain if the eggs are removed three at a time. If the eggs are removed four, five, or six at a time, the three, four, and five eggs will remain, respectively. However, no eggs will be left over if they are taken out seven at a time.
Find the smallest number of eggs that could be in the basket.

Correct answer:

n =  119

Step-by-step explanation:

$$\begin{gathered} n=2x+1 \\ n=3y+2 \\ n=4z+3 \\ n=5w+4 \\ n=6q+5 \\ n=7k \\ 2...\text{ prime number} \\ 3...\text{ prime number} \\ 4=2^2 \\ 5...\text{ prime number} \\ 6=2\cdot 3 \\ LCM(2,3,4,5,6)=2^2\cdot 3\cdot 5=60 \\ {z}_1=LCM(2,3,4,5,6)=60 \\ {t}_1=(1\cdot z_1-1)/7=(1\cdot 60-1)/7=\frac{59}{7}=8\frac{3}{7}\doteq 8.4286 \\ {t}_2=(2\cdot z_1-1)/7=(2\cdot 60-1)/7=17 \\ {t}_3=(3\cdot z_1-1)/7=(3\cdot 60-1)/7=\frac{179}{7}=25\frac{4}{7}\doteq 25.5714 \\ {t}_4=(4\cdot z_1-1)/7=(4\cdot 60-1)/7=\frac{239}{7}=34\frac{1}{7}\doteq 34.1429 \\ {t}_5=(5\cdot z_1-1)/7=(5\cdot 60-1)/7=\frac{299}{7}=42\frac{5}{7}\doteq 42.7143 \\ \dots \\ n=2\cdot z_1-1=2\cdot 60-1=119 \\ \text{ Verifying Solution: } \\ 119=2\times 59+1 \\ 119=3\times 39+2 \\ 119=4\times 29+3 \\ 119=5\times 23+4 \\ 119=6\times 19+5 \\ 119=7\times 17 \end{gathered}$$

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