What is the <strong>LEAST</strong> value of n if 194480+n=m^4 where m and n are natural numbers?

Explanation

We need the smallest fourth power greater than 194480. Since 20^4 = 160000 and 21^4 = 194481, the smallest value is 194481. Thus, 194480 + n = 194481 \implies n=1.

Related questions on Arithmetic

• What is the remainder when (17^{25} + 19^{25}) is divided by 18?
• A bottle contains spirit and water in the ratio 1:4 and another identical bottle contains spirit and water in the ratio 4:1. In what rat...
• Let P = 5^5 \times 15^{15} \times 25^{25} \times 35^{35} and Q = 10^{10} \times 20^{20} \times 30^{30} \times 40^{40}. What is the numbe...
• Two students X and Y appeared in a test. The score of X is 20 more than that of Y. If the score of X is 75% of the sum of the scores of X an...
• Question: The product of a natural number N and the number M written by the same digits of N in the reverse order is 252. What is the number...