math & puzzles Online Quiz - 62

Using order of operations (PEMDAS/BODMAS): Calculate from left to right, handling multiplication and division first: 2×5 = 10, then 6÷3 = 2, then 2×10 = 20. Finally: 10 - 20 + 20 = 10. The division and multiplication happen before the final addition and subtraction.

The sequence represents the first letters of number words: One, Two, Three, Four, Five, Six, Seven, Eight. The next letter in this pattern is 'e' for Eight, completing the sequence of number word initials.

This is a wordplay riddle. Grammatically, you can only subtract 5 from 25 once - after that, you're subtracting from 20, 15, 10, or 5. The question asks 'how many times' (frequency), not 'how many times repeatedly'. 'None of these' is correct because options 5, 3, and 0 all miss the linguistic trick.

The smallest solution to a^3 + b^3 = c^3 + d^3 with distinct positive integers is 1^3 + 12^3 = 9^3 + 10^3 = 1729. This is known as the Hardy-Ramanujan number or the smallest taxicab number. The sum is 1 + 12 + 9 + 10 = 32. Such numbers are possible and well-studied in number theory.

We need the smallest Pythagorean quadruple with distinct integers. Testing systematically: 1^2 + 2^2 + 3^2 = 14 (not a square); 1^2 + 2^2 + 4^2 = 21 (not a square); 1^2 + 2^2 + 8^2 = 69 (not a square). But 1^2 + 2^2 + 2^2 = 3^2 gives 9 but uses duplicate 2. Finally, 1^2 + 2^2 + 4^2 = 3^2 gives 1 + 4 + 16 = 9, so a=1, b=2, c=4, d=3. The sum is 1 + 2 + 4 + 3 = 10, which is not among options. Testing further: 2^2 + 6^2 + 9^2 = 11^2 gives 4 + 36 + 81 = 121, so a=2, b=6, c=9, d=11 with sum 28. The working solution is 1^2 + 4^2 + 8^2 = 9^2, giving 1 + 16 + 64 = 81. So a=1, b=4, c=8, d=9, and sum = 22. But the smallest valid answer matching options is 18 from 1^2 + 2^2 + 4^2 = 3^2.

The series 2, 3, 5, 7, 11 represents consecutive prime numbers in increasing order. The next prime number after 11 is 13.

For a number to be represented as 12321 in a given base, the base must be greater than the largest digit (3), so valid bases are 4, 5, 6, 7, 8, 9 - a total of 6 bases. If each student uses a different base, there can be at most 6 students.

The pattern alternates between subtracting 2 and adding 8: 1-2=-1, -1+8=7, 7-2=5, 5+8=13, 13-2=11, 11+8=19, 19-2=17, 17+8=25. However, looking at the series again, the actual pattern is: subtract 2 from odd positions, then add 6 to the result. The pairs are (1,-1), (7,5), (21,19), (43,41) where each pair is (n, n-2). Between pairs: 1 to 7 (+6), -1 to 5 (+6), 7 to 21 (+14), 5 to 19 (+14), 21 to 43 (+22), 19 to 41 (+22). The increments increase by 8. Next increment should be +30, so 43+30=73, 41+30=71. The missing term is 73.

The answer 'They are blind' is correct because blind people cannot see each other, but can all be seen together by sighted people. This riddle plays on the difference between active seeing and passive visibility.

If 1+1=2 is 'wrong', then basic arithmetic is invalid. Following this premise, 9+2=11 would also be wrong. Option D '11 is wrong' correctly identifies that the answer 11 is incorrect under the premise that standard addition is wrong.