If 1 is added to each odd digit and 2 is subtracted from each even digit in the number 6238571, how many digits will appear more than once in the new number thus formed?
2
Treat each digit of 6238571 separately: an odd digit (1, 3, 5, 7, 9) gets +1 and an even digit (0, 2, 4, 6, 8) gets –2.
6 (even) → 6 – 2 = 4; 2 (even) → 2 – 2 = 0; 3 (odd) → 3 + 1 = 4; 8 (even) → 8 – 2 = 6; 5 (odd) → 5 + 1 = 6; 7 (odd) → 7 + 1 = 8; 1 (odd) → 1 + 1 = 2.
So the new number is 4046682. Counting each digit: 4 appears twice, 6 appears twice, while 0, 8 and 2 appear once each.
Therefore the digits that appear more than once are 4 and 6, i.e. 2 digits.
Check: the total count 2 (for 4) + 1 (for 0) + 2 (for 6) + 1 (for 8) + 1 (for 2) = 7 matches the seven digits of the original number, so no digit was missed.
Hence, 2.
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Important Questions from Operations on Place Value
In a certain code language, TOGETHER is coded as 119 and MASTER is coded as 87. How will REVERSE be coded in that language?
In a certain code, if BRDKE is written as AQCJD, then SLEEP will be written in that code as:
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In a code language, 'SPACE' is written as '5-91' and 'ROTTEN' is written as '6-70'. How will 'MEAN' be written in that language?
If in a coding system, ORATORY is coded as 116 and RAVENOUS is coded as 119, then how will IRIDOLOGY be coded in the same coding system?
