What is the number of integers from 1 to 1000 (inclusive) that are divisible by neither 11 nor by 35?
- 884
- 890
- 892
- 910
- 945
To count the number of integers from 1 to ![[m]N[/m]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tOZT7Ws-C9IP-IVI_qT_pqpMLFWZ5IcS_frYKoGHc9FjfspDoLE3FlM8xIpS856HcUsTOfxD7uOEilsctW3OhjB9vdlg9823I2wMLj=s0-d)
(inclusive) that are divisible by ![[m]x[/m]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_shTdXkjqB4LlpRL-oPOtxX_ePRRfCXkyJ2EEd8SLQ3AH0q30twDoGD0MM4O3idQP23E9P5BNkrnYAtRE-TumngZV8hxdDBm5Rka-ov=s0-d)
, find the value of ![[m]\frac{N}{x}[/m]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t627jTHfLeSKR0H0_qbc0cw_nVsZJjIYiDHEh8qKB_BKriTG2Vr3BIp64hNl0u9BiNPs-KeqCxeOx0PEbJs06N2srga40zPda4cO6FROkTZQ9Qfu61kpRQEq028xVdo4s=s0-d)
. Use only the integer part of the resulting number. Based on the formula, the number of integers divisible by 11 is ![[m]\frac{1000}{11} = 90.9 = 90[/m]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_snIPAeL5Wrt9hBfKn1eUB9NR3xS4ywbwesKDuK729FngYX_Z2DKIPtTmLjAiqe51d0I6I2R1nHCo5ebvsinwKVXgfJ9tpQdC_xYQWtS2AzV4UgWGm5aSLgsf095tcjz2Vna1IXKQ7I8I-03pXOd_9loGVrmA=s0-d)
(even though the result could be rounded off to 91, use 90). In the same way, the number of the integers divisible by 35 is ![[m]\frac{1000}{35} = 28.57 = 28[/m]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sbrR_QWfpKKaG9NTlcYdUXgDk55eXW2DnGAdzLeSz9kQQPKlypB1S8AvBVFrQj6EthBbjp1CQkAehzkD6DYixyjWb5r5ONdtl0VMUb3vaeJWwuif7D4A_CKMpa8qUzOg-7cb3dQa-7P2xdt4KnLy4O1aayLg=s0-d)
.
Subtract the number of integers that are divisible by both 11 and 35, so that they are not counted twice.
Therefore,![[m]\frac{1000}{11} + \frac{1000}{35} - \frac{1000}{11*35} = 90 + 28 - 2 = 116[/m]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uLLNSLF-fi85b9eoPltZwMGK4RZ2ys2xgwtWmgpu2Rz1XRPbaVzuRr_mYsWXvkI_J0ZZ-SB2aZnZlmK-i9QlvlPJEIFrOKDg_e1u8-M_c6_NEfmJ8fYYXjRtktJCDmKgE24LIdajS-4j5rDZjO722EM1EGU2t8j8PBrqBxiryZVnnbo7YRH640dxT5tHjPF5faJ9hFCxGiBsKPVk02S8VxeSI1ADXuhxivDujaZsnK_r2a-ePKyzC1_u4Fug=s0-d)
.
![[m]1000-116 = 884[/m]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uUY9lGvHgYz84mvPIUksETDhGo6pnTJc43GgR5WKpOCr-UR49h0Qkmsb3Ms6rjowmggD6oRU5Q1dgpvaGU_iQooJ42PL7kZxn_ygaGdtW1PqFoyLD-Yv4kXrE=s0-d)
.
(inclusive) that are divisible by , find the value of . Use only the integer part of the resulting number. Based on the formula, the number of integers divisible by 11 is (even though the result could be rounded off to 91, use 90). In the same way, the number of the integers divisible by 35 is .Subtract the number of integers that are divisible by both 11 and 35, so that they are not counted twice.
Therefore,
. .The correct answer is A.
My answer was right. Wanted to keep this here to understand why they've used 1000/ 11*35. What I did was calculate 33*11 and 33*11*2 and saw that both are under 1000 and then counted 2. I should know this other method too.
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