Sure, let's solve this step by step:

Step 1: We know that (a^b + a) is always divisible by (a + 1) when b is odd. This is because (a^b + a) can be written as a*(a^(b-1) + 1), which is divisible by (a + 1) when b is odd.

Step 2: In this case, a = 67 and b = 67. Since b is odd, we can say that (67^67 + 67) is divisible by (67 + 1) = 68.

Step 3: If a number is divisible by another number, the remainder is always 0.

So, the remainder when (67^67 + 67) is divided by 68 is 0.

Question

Sure, let's solve this step by step:

Step 1: We know that (a^b + a) is always divisible by (a + 1) when b is odd. This is because (a^b + a) can be written as a*(a^(b-1) + 1), which is divisible by (a + 1) when b is odd.

Step 2: In this case, a = 67 and b = 67. Since b is odd, we can say that (67^67 + 67) is divisible by (67 + 1) = 68.

Step 3: If a number is divisible by another number, the remainder is always 0.

So, the remainder when (67^67 + 67) is divided by 68 is 0.

Knowee AI · Accepted Answer

Sure, let's solve this step by step:

Step 1: We know that (a^b + a) is always divisible by (a + 1) when b is odd. This is because (a^b + a) can be written as a*(a^(b-1) + 1), which is divisible by (a + 1) when b is odd.

Step 2: In this case, a = 67 and b = 67. Since b is odd, we can say that (67^67 + 67) is divisible by (67 + 1) = 68.

Step 3: If a number is divisible by another number, the remainder is always 0.

So, the remainder when (67^67 + 67) is divided by 68 is 0.

If (67^{67}+67) is divided by 68, the remainder is:

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