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Question

Chứng tỏ rằng A = 24 + 25 + 26 + 27 + 28 + 29 chia hết cho 3

Full Moon · Accepted Answer

Ta có:$A=2^4+2^5+2^6+2^7+2^8+2^9$$=\left(2^4+2^5\right)+\left(2^6+2^7\right)+\left(2^8+2^9\right)$$=2^4\left(1+2\right)+2^6\left(1+2\right)+2^8\left(1+2\right)$$=3\left(2^4+2^6+2^8\right)⋮3$Vậy A chia hết cho 3Câu trả lời: Ta có:$A=2^4+2^5+2^6+2^7+2^8+2^9$$=\left(2^4+2^5\right)+\left(2^6+2^7\right)+\left(2^8+2^9\right)$$=2^4\left(1+2\right)+2^6\left(1+2\right)+2^8\left(1+2\right)$$=3\left(2^4+2^6+2^8\right)⋮3$Vậy A chia hết cho 3

Tẫn · Answer

Ta có:$A=2^4+2^5+2^6+2^7+2^8+2^9$$=(2^4+2^5)+(2^6+2^7)+(2^8+2^9)$$=\left[2^4.\left(1+2\right)\right]+\left[2^6.\left(1+2\right)\right]+\left[2^8.\left(1+2\right)\right]$$=2^4.3+2^6.3+2^8.3$$=3.\left(2^4+2^6+2^8\right)$$\Rightarrow A⋮3$Câu trả lời: Ta có:$A=2^4+2^5+2^6+2^7+2^8+2^9$$=(2^4+2^5)+(2^6+2^7)+(2^8+2^9)$$=\left[2^4.\left(1+2\right)\right]+\left[2^6.\left(1+2\right)\right]+\left[2^8.\left(1+2\right)\right]$$=2^4.3+2^6.3+2^8.3$$=3.\left(2^4+2^6+2^8\right)$$\Rightarrow A⋮3$

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