EPFO EO/AO 2023 General Ability Test · Question 71 of 120

Correct — D, (d) x = 4, y = 4. Take 88 apart first: 88 = 8 × 11, and 8 and 11 are coprime, so a number is divisible by 88 exactly when it is divisible by both 8 and by 11. Those two tests can be applied independently, and between them they pin x and y down completely without any trial of the options. The test for 11. Alternate the signs of the digits and add. For 7 8 9 4 5 9 x y that is 7 − 8 + 9 − 4 + 5 − 9 + x − y. The six known digits cancel exactly: 7 − 8 = −1, +9 = 8, −4 = 4, +5 = 9, −9 = 0. So the alternating sum is simply x − y, and divisibility by 11 requires x − y to be a multiple of 11. Since x and y are single digits, x − y lies between −9 and 9, and the only multiple of 11 in that range is 0. Therefore x = y. That one step disposes of options (a) and (c) immediately, since 1 ≠ 2 and 3 ≠ 6. The test for 8. A number is divisible by 8 when its last three digits are. Here they are 9, x, y — the number 900 + 10x + y. Now 900 = 8 × 112 + 4, so 900 leaves remainder 4, and 10x leaves the same remainder as 2x. The condition becomes 4 + 2x + y ≡ 0 (mod 8), that is 2x + y ≡ 4 (mod 8). Put the two together. With y = x the condition is 3x ≡ 4 (mod 8). Running x through the digits: 3×0 = 0, 3×1 = 3, 3×2 = 6, 3×3 = 9 ≡ 1, 3×4 = 12 ≡ 4. That works, and the next solution would be x = 4 + 8 = 12, which is not a digit. So x = 4 is the only possibility and y = 4 with it. Check the answer against the original number: 78945944. Its last three digits, 944, give 944 ÷ 8 = 118 exactly, and its alternating digit sum is 7 − 8 + 9 − 4 + 5 − 9 + 4 − 4 = 0, which is divisible by 11. So 78945944 is divisible by 88 — indeed 78945944 ÷ 88 = 897113. The solution is unique, which is why the paper can ask for 'the possible values' in the plural and still key a single option.

• (a)x = 1, y = 2 — Fails the test for 11. The alternating digit sum reduces to x − y, which here is 1 − 2 = −1, not a multiple of 11, so 78945912 is not divisible by 11 and cannot be divisible by 88. It does pass the test for 8 — 912 ÷ 8 = 114 — which is exactly what makes it dangerous: a candidate who checks only the easier of the two tests will accept it. Both factors of 88 have to be checked, and the eleven test is the one that does the real work here.
• (b)x = 2, y = 2 — This one passes the test for 11, since x = y makes the alternating sum zero, and fails the test for 8. The last three digits are 922, and 922 ÷ 8 = 115.25. Put algebraically, the condition 3x ≡ 4 (mod 8) gives 3 × 2 = 6, not 4. So 78945922 is divisible by 11 but not by 8, and therefore not by 88. It is the mirror image of option (a) and catches the opposite half-solution.
• (c)x = 3, y = 6 — Fails the test for 11: x − y = 3 − 6 = −3, which is not a multiple of 11. Like option (a) it survives the test for 8, since 936 ÷ 8 = 117, so 78945936 is divisible by 8 and even by 4 and by 9, but not by 11 and so not by 88. Notice that the three wrong options between them cover every partial approach — two pass only the eight test, one passes only the eleven test — which is the examiner's way of making sure the answer cannot be reached by doing half the work.

Divisibility questions with unknown digits are solved by factorising the divisor into coprime parts and testing each part separately, because a number is divisible by a product of coprime factors exactly when it is divisible by each of them. 88 = 8 × 11 is the natural split. The rule for 8 is that the last three digits must form a multiple of 8, which follows from 1000 being a multiple of 8, so everything to the left of the hundreds place contributes nothing. The rule for 11 is the alternating digit sum, which follows from 10 ≡ −1 (mod 11): each power of ten is alternately +1 and −1, so the number is congruent to the alternating sum of its digits. Written as congruences rather than as tests, both rules turn a search over a hundred digit-pairs into two linear conditions in x and y, and two linear conditions in two unknowns usually pin the answer down exactly — as they do here. The same technique handles 44 = 4 × 11, 72 = 8 × 9, 99 = 9 × 11 and 132 = 4 × 3 × 11. What it never handles is a divisor split into factors that share a factor: 88 is not 4 × 22 for this purpose, because 4 and 22 have 2 in common and passing both tests would not guarantee divisibility by 88.

The quantitative block of the EO/AO paper favours number theory that can be done without a calculator in under a minute, and unknown-digit divisibility is the archetype. Because all four options are supplied, a candidate can brute-force this question by long division; the reason to solve it properly is time, since the algebraic route takes two lines and settles all four options at once. The habit rewarded is factorising the divisor before touching the digits.

• 88 = 8 × 11, and 8 and 11 are coprime, so divisibility by 88 is equivalent to divisibility by 8 and by 11 together.
• Divisibility by 8: the number formed by the last three digits must be divisible by 8, because 1000 is a multiple of 8.
• Divisibility by 11: the alternating sum of the digits must be a multiple of 11, because 10 ≡ −1 (mod 11).
• For 789459xy the alternating sum of the six known digits is zero, so the eleven-condition reduces to x − y ≡ 0 (mod 11), and for single digits that forces x = y.
• The eight-condition on the last three digits 9xy is 900 + 10x + y ≡ 0 (mod 8), which simplifies to 2x + y ≡ 4 (mod 8) since 900 ≡ 4 and 10x ≡ 2x.
• Combining them gives 3x ≡ 4 (mod 8), whose only single-digit solution is x = 4; hence x = y = 4 and the number is 78945944 = 88 × 897113.
• The method fails if the divisor is split into factors sharing a common factor — 88 must be split as 8 × 11, never as 4 × 22.

• Testing only one factor. Two of the three wrong options here are divisible by 8, and one is divisible by 11.
• Splitting 88 as 4 × 22. The factors must be coprime for the two tests to be jointly sufficient.
• Applying the eight rule to the last two digits. That is the rule for 4; for 8 it is the last three.
• Getting the alternating sum's direction wrong. The sign pattern must be consistent, and since only a multiple of 11 is being sought, starting from either end gives the same verdict.

EO/AO quantitative items on number properties are short and mechanical: an unknown digit to be found, an LCM or HCF to be combined, a unit digit of a large power, a remainder. All of them reward knowing the standard divisibility tests cold and being willing to write one congruence instead of testing four options by division.

No directly related past PYQ was found.

• practice — not a real PYQ

  The five-digit number 485x6 is divisible by 44. What is the value of x ?

  • (a)3
  • (b)5
  • (c)7
  • (d)9

  Answer(c) 7

• practice — not a real PYQ

  For a number to be divisible by 8, which one of the following must hold ?

  • (a)The sum of its digits is divisible by 8
  • (b)The number formed by its last two digits is divisible by 8
  • (c)The number formed by its last three digits is divisible by 8
  • (d)The alternating sum of its digits is divisible by 8

  Answer(c) The number formed by its last three digits is divisible by 8