Last digit of a power sum: keep each cycle position

The powers in 7²⁰²⁶ + 3²⁰²⁵ are large, but the question is small: what is the final decimal digit of their sum? You do not need to write either full power. In this original integer problem, track the ones digit of each term, find its position in a repeating cycle, and then keep the ones digit of the combined result. The answer is 2.

For the Last digit of a power sum, retain the exact expression beside the work. The first base is 7 with positive exponent 2026; the second base is 3 with positive exponent 2025. The operation between them is addition. These are invented mathematical inputs, not dates, prices, measurements or specifications for a writing board. Changing a base, exponent or operation requires checking the corresponding calculation again.

Keep only the information the question requests

A nonnegative integer can be written as ten times a whole number plus its ones digit. For example, 49 is ten times 4 plus 9. When that number is multiplied by 7, the ten-times part still contributes a multiple of ten. Only the product of its ones digit and 7 is needed to determine the next ones digit.

This gives a simple recurrence: multiply the current ones digit by the base, then keep the ones digit of that product. You are not claiming that the full numbers are equal to their last digits. You are deliberately retaining one piece of information that is sufficient for the requested output. The other digits remain unknown and are not needed here.

The same distinction matters at the final addition. If one term is ten times an integer plus 9, and the other is ten times an integer plus 3, their sum is a multiple of ten plus 12. Its ones digit is therefore the ones digit of 12. A carry into the tens place does not change that final digit.

Build the cycle for powers of seven

Begin with the first four positive powers of 7. They are 7, 49, 343 and 2401. Their ones digits are 7, 9, 3 and 1. Write these under positions one, two, three and four. The position labels refer to the exponent, rather than the number of digits in the full power.

Check the transitions using only the retained digits. Seven times 7 gives 49, so a current 7 is followed by 9. Nine times 7 gives 63, so 9 is followed by 3. Three times 7 gives 21, so 3 is followed by 1. Finally, one times 7 gives 7, returning to the starting digit.

The return is enough to justify continuing the cycle because the transition rule is fixed. Once the same current digit occurs again, multiplying it by the same base gives the same next digit. Thus the positive-power sequence repeats as 7, 9, 3, 1, then 7, 9, 3, 1 again. You have established a rule for these powers, rather than guessed a pattern from a few large values.

Attach the first exponent to its cycle position

Divide 2026 into complete groups of four and a remainder: 2026 = 4 × 506 + 2. The 506 complete cycles return through the same four positions. The remaining two positions select the second entry in the cycle for powers of 7. That entry is 9, so 7²⁰²⁶ has ones digit 9.

Keep the base name beside this result. A useful note says “base 7; exponent remainder 2; selected digit 9.” A loose remainder 2 is not itself the last digit. It tells you which entry to read from the identified cycle. Confusing the index with the selected value would replace the correct 9 with 2.

Nor should the exponent remainder be read as a count of extra digits to write. The full power has not been calculated. The remainder only locates a position in the proved four-step recurrence. That is all the information the current question requires from the first term.

Build and use the separate cycle for three

The first four positive powers of 3 are 3, 9, 27 and 81. Their ones digits are 3, 9, 7 and 1. Check the transitions: 3 times 3 gives 9; 9 times 3 gives 27; 7 times 3 gives 21; and 1 times 3 gives 3. The same starting digit has returned under a fixed transition rule.

The second cycle therefore repeats as 3, 9, 7, 1. It has the same length as the first cycle, but its entries are not in the same order. Keep it on a separate line. Reusing the first cycle merely because both have four positions would ignore the base that determines each transition.

For the second exponent, write 2025 = 4 × 506 + 1. The remaining one selects the first position of the base-3 cycle, whose digit is 3. Thus 3²⁰²⁵ has ones digit 3. The two term records are now complete: the first contributes a last digit of 9 and the second a last digit of 3.

Combine the selected digits and finish the requested task

Add 9 and 3 to get 12. The final decimal digit of 12 is 2, so the original sum has final digit 2. Do not stop at 12: the problem requested one final digit, rather than the unreduced sum of the two retained digits. The tens contribution is carried into a place the question did not ask you to report.

You can express the reasoning without knowing either huge integer. Write the first power as 10A + 9 and the second as 10B + 3, where A and B are nonnegative integers representing their higher-place parts. Their sum is 10(A + B + 1) + 2. This directly shows why the carry leaves final digit 2.

That equation does not identify A, B or the full sum. It certifies the requested final digit under the stated input. Do not use the answer 2 as a claim about the first digit, number of digits or approximate magnitude of the sum. Those would be different questions requiring additional information and work.

Check the zero-remainder case before reusing the method

A common indexing mistake appears when an exponent is divisible by four. As a separate control, 2024 = 4 × 506 has remainder zero. In either of these positive-power cycles, that selects the fourth position, whose digit is 1. It does not select a digit zero or a nonexistent “zeroth” entry on a one-through-four list.

Check a small example alongside that convention. The exponent 4 has remainder zero upon division by four. Seven to the fourth is 2401, and three to the fourth is 81; both end in 1. Their visible values agree with the fourth-position rule. This control helps detect a mistaken conversion from a remainder to a cycle index.

Keep the present explanation within its stated conditions: the two particular bases, positive whole-number exponents, and a decimal ones-digit question. It does not claim that every base has a cycle of length four. Before using this approach on another base, establish that base's actual transition sequence and the cycle positions relevant to the new task.

Repair the record at the step that changed

If a worksheet gives 3, inspect whether it added the exponent remainders instead of the selected digits. Here the remainders 2 and 1 locate positions; they are not replacements for the powers' ones digits. Return to the separate base-labelled cycles and read their entries before combining anything.

If the record gives 12 as its final answer, retain that intermediate sum and add the missing final selection. The ones digit is 2. If the first term was assigned 3, compare its recorded remainder with its own cycle: position two in 7, 9, 3, 1 is 9. A correction should identify the mistaken lookup rather than erase the whole calculation.

For a reproducible explanation, retain the two cycles, the two exponent divisions and the final addition. Someone checking the work can verify those short lines independently. Expanding a large power would add a long integer to copy without improving the central explanation of why a four-position lookup solves this specific problem.

Separate nearby arithmetic ideas

The tens and ones abacus exercise assigns place weights and exchanges ten ones for one ten. This problem uses decimal place value to retain a final digit, but it teaches no bead layout, active-side convention or physical carrying sequence. Its main task is selecting exponent-cycle positions.

The repeating decimal to fraction tutorial aligns stated infinite decimal tails and checks a fraction through long division. Here the repeated object is a finite sequence of possible ones digits under multiplication. There is no repeating-decimal input, tail cancellation or fraction conversion to perform.

Read the board offers as drawing-surface context

The Quartet EMA408 whiteboard listing describes a white melamine writing surface and satin-finish aluminum frame, with stated 96-by-48-inch dimensions. Its five gallery images include a whole board, tray and mounting-hardware inclusion graphics, and an erasing demonstration. Those images do not independently establish the material, measured size, erasing performance or supplied marker inventory.

The Quartet PPA406 whiteboard listing describes a porcelain magnetic board with an aluminum frame and stated 72-by-48 dimensions. Its four gallery images include the complete board, a tray detail and illustrative attachment and erasing scenes. No magnetic holding test, material certification, scratch resistance result or supplied tool inventory is established here.

A suitable drawing surface can keep the two cycles on separate lines. That is presentation space for the authored problem, not a finding that either board performs arithmetic or includes the tools in every demonstration. Retain the exact mathematical inputs and short lookup records so the explanation remains readable independently of a purchase.

Keep any cryptocurrency purchase attached to its own evidence

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