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ATQ,
For divisibility by 6, the number must be divisible by both 2 and 3.
Sum of digits = 8 + 3 + 2 + 1 + 9 + y = 23 + y
For divisibility by 3, (23 + y) must be divisible by 3.
Also, y should be even (for divisibility by 2) and y ≠ 0 or 5 (to avoid divisibility by 5).
Even digits: 0, 2, 4, 6, 8
Trying values:
23 + 4 = 27 → divisible by 3 ✅ and 4 is even ✅ → y = 4
So, (2y + 5) = 2×4 + 5 = 13
(71.78 × 59.88 ÷ 17.92 ÷ 120.22 × 161.72)% of 32 = ?
13.99% of 299.99 ÷ 7.17 = ? ÷ 16.15
8.15 of 124.95 ÷ 40.13 + 249.84 X 14.18 - √325 X 149.87 = ? X 10.85
(22.93 × 11.92) + (17.78 ÷ 3.01) - (14.88 × 5.01) = ?
4 √1295.98 × 1339.42 + ? = 73 × 26.01
? × 32.91 – 847.95 ÷ √16.4 – 13.982 = √24.7 × 24.04
14.2% of 7200 + 2.8% of 6400 =?
?% of (95.31 ÷ 18.97 × 70.011) = 174.98
20.11% of 119.99 + √80.97 ÷ 3.02 = ?
[(5/9 of 719.87) + (59.73% of 450.31)] ÷ (√168.79 - 3/4 of 63.94) = ?