Explanation
Solving
Let the number of coins be x. Based on the given conditions, we can write the following congruences:
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x≡1(mod3) (which is x=3k+1, also can be seen as x=3k−2)
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x≡2(mod4) (which is x=4k+2, also can be seen as x=4k−2)
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x≡3(mod5) (which is x=5k+3, also can be seen as x=5k−2)
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x≡4(mod6) (which is x=6k+4, also can be seen as x=6k−2)
Observation:
In each case, the remainder is 2 less than the divisor. This means if we add 2 to the number of coins, the result will be exactly divisible by 3,4,5, and 6.
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Find the LCM:
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General Form:
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Since x < 100:
Substitute n=1:
Checking the result:
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58=(3×19)+1 (Satisfies i)
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58=(4×14)+2 (Satisfies ii)
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58=(5×11)+3 (Satisfies iii)
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58=(6×9)+4 (Satisfies iv)
Correct Option: 1
