A number theory problem by أحمد الحلاق

$70! + 71! + 72! = 70! × (1 + 71 + 71 × 72) = 70! × 72^2 = 70! × (2^6 × 3^4)$

Now, if we want to determine the highest power of 3, which is a factor of 70! , then we can consider the following:

The number of integers between 1 and 70 (inclusive), which are multiples of:

• $3 : \bigl \lfloor { \frac {70}{3}} \bigr \rfloor = 23$

• $3^2 = 9: \bigl \lfloor { \frac {70}{9}} \bigr \rfloor = 7$

• $3^3 = 27: \bigl \lfloor { \frac {70}{27}} \bigr \rfloor = 2$

Therefore, 70! can be written as:

$70! = 3^{23 + 7 + 2} × A = 3^{32} × A \text {, where } A \in \mathbb {N} , 3 \nmid A$

$70! + 71! + 72! = 70! × (2^6 × 3^4) = 3^{32} × A × (2^6 × 3^4) = 2^6 × 3^{4 + 32} × A = 2^6 × 3^{ \boxed {36}} × A$